16
SCH3 U R.H. KING ACADEMY Moles & Stoichiometry WORKSHEET Name: W1 Avogadro’s Number, Moles and Molar Mass For a scientist to carry out experiments involving only 1, 2, 3, a dozen or even a million atoms or molecules is not feasible. The volume that this many atoms or molecules would occupy is negligible. Reasonable masses of elements for scientists to work with are those that are based on the average atomic mass for an element. It was determined by the Italian scientist, Avogadro, that the number of atoms in exactly 12.000 g of the carbon-12 isotope was 6.02 x 10 23 . In fact a sample of any element equal to its average atomic mass contains 6.02 x 10 23 atoms of that element. Scientists refer to this amount of any entity as a “mole” and use this term in the same manner that a baker would use the unit of a dozen. The mole (symbol “n”, unit mol) is described as the number of particles in exactly 12.000 grams of C ! !" . The number 6.02 x 10 23 is called Avogadro’s number and is represented by the symbol “N A ”. Avogadro’s number is so big it is in comprehensible, below are some examples to help grasp the magnitude of this number. 1. If 1 million ants live in an ant hill 1 m 2 , 1 mole of ants would occupy enough ant hills to cover Earth 1000 times. 2. To travel a distance of 1 mole of millimetres at the speed of light (3.0 x10 8 m/s) it would take 63 654 years. The mass of 1 mole of any substance is referred to as it Molar Mass (symbol “MM”, unit g/mol) and for an element is equal to the atomic mass on the periodic table in g/mol (for a specific isotope, the mass number is used). The molar mass of a compound is equal to the algebraic sum of the atomic masses for each element in the formula (multiply each element’s mass by the number of that element in the formula, then sum). Calculations Using the Mole There are 2 formulas for calculating the number of moles or required quantities (number of particles or mass). Formula 1: Converting number of particles to moles n = # of particles N ! where, n= # of moles N A = Avogadro’s # Formula 2: Converting mass into moles n = m MM where, m = mass in grams MM = molar mass in g/mol Examples: 1. Determine the moles of Ca in 7.56x10 24 atoms? n = # of particles N ! n = 7.56×10 !" 6.02×10 !" n = 12.6 mol 2. How many CO 2 molecules are in 3.25 mol? n = # of particles N ! # of particles = nN ! # of particles = (3.25 mol)(6.02× 10 !" molecules mol ) # of particles = 1.96×10 !" molecules 3. How many moles are in 12.75 g of sodium? n = m MM n = 12.75 g 22.99 g/mol n = 0.5546 mol 4. Calculate the mass of 2.50 mol of Mg 3 (PO 4 ) 2 . MM = 3 Mg + 2 P + 8 O MM = 3 24.31 g mol + 2 30.97 g mol + 8(16.00 g mol ) MM = 262.87 g/mol n = m MM m = n×MM m = 2.50 mol 262.87 g mol m = 657 g

Avogadro’s Number, Moles and Molar Masssch3uking.wikispaces.com/file/view/MOLES+&+STOICHIOMETRY+WO… · Avogadro’s Number, Moles and Molar Mass For a scientist to carry out experiments

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SCH3  U-­‐  R.H.  KING  ACADEMY                                                    Moles  &  Stoichiometry  WORKSHEET   Name:        

W1    

Avogadro’s Number, Moles and Molar Mass

For a scientist to carry out experiments involving only 1, 2, 3, a dozen or even a million atoms or molecules is not feasible. The volume that this many atoms or molecules would occupy is negligible. Reasonable masses of elements for scientists to work with are those that are based on the average atomic mass for an element. It was determined by the Italian scientist, Avogadro, that the number of atoms in exactly 12.000 g of the carbon-12 isotope was 6.02 x 1023. In fact a sample of any element equal to its average atomic mass contains 6.02 x 1023 atoms of that element. Scientists refer to this amount of any entity as a “mole” and use this term in the same manner that a baker would use the unit of a dozen. The mole (symbol “n”, unit mol) is described as the number of particles in exactly 12.000 grams of C!!" . The number 6.02 x 1023 is called Avogadro’s number and is represented by the symbol “NA”. Avogadro’s number is so big it is in comprehensible, below are some examples to help grasp the magnitude of this number.

1. If 1 million ants live in an ant hill 1 m2, 1 mole of ants would occupy enough ant hills to cover Earth 1000 times. 2. To travel a distance of 1 mole of millimetres at the speed of light (3.0 x108 m/s) it would take 63 654 years.

The mass of 1 mole of any substance is referred to as it Molar Mass (symbol “MM”, unit g/mol) and for an element is equal to the atomic mass on the periodic table in g/mol (for a specific isotope, the mass number is used). The molar mass of a compound is equal to the algebraic sum of the atomic masses for each element in the formula (multiply each element’s mass by the number of that element in the formula, then sum).

Calculations Using the Mole

There are 2 formulas for calculating the number of moles or required quantities (number of particles or mass).

Formula 1: Converting number of particles to moles

n =#  of  particles

N!

where, n= # of moles NA= Avogadro’s #

Formula 2: Converting mass into moles

n =mMM

where, m = mass in grams MM = molar mass in g/mol

Examples:

1. Determine the moles of Ca in 7.56x1024 atoms?

n =#  of  particles

N!

n =7.56×10!"

6.02×10!"

n = 12.6  mol

2. How many CO2 molecules are in 3.25 mol?

n =#  of  particles

N!

#  of  particles = nN!

#  of  particles = (3.25  mol)(6.02×10!"  molecules

mol)

#  of  particles = 1.96×10!"molecules

3. How many moles are in 12.75 g of sodium?

n =mMM

n =12.75  g

22.99  g/mol

n = 0.5546  mol

4. Calculate the mass of 2.50 mol of Mg3(PO4)2. MM = 3  Mg + 2  P + 8  O

MM = 3 24.31gmol

+ 2 30.97gmol

+ 8(16.00gmol

)

MM = 262.87 g/mol

n =mMM

m = n×MM

m = 2.50  mol 262.87gmol

m = 657  g

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SCH3  U-­‐  R.H.  KING  ACADEMY                                                    Moles  &  Stoichiometry  WORKSHEET   Name:        

W2    

Exercise: Complete the following on a separate piece of paper and express the answer to the correct number of significant digits.

1. State the mass of 1 mole of the following in g/mol. a) B!!" b) nitrogen c) Na!!

!!

d) S e) I!"

!"# f) thallium

2. Calculate the molar mass of the following compounds. a) HBr b) NO2 c) Cu d) Mg(OH)2

e) Fe2(SO4)3 f) C2H6O g) C12H26 h) Sn3(PO3)4

3. Determine the number of moles in the following. a) 6.02 x 1023 atoms of N b) 4.52 x 1023 electrons

c) 3.011 x 1024 molecules of H2O d) 4.25 x 1025 Ca2+ ions

4. Calculate the number of particles in the following. a) 1.50 mol of Ba atoms b) 3 mol of C atoms

c) 0.75 mol of H2O molecules d) 2.67 mol of NaOH molecules

5. Determine the number of atoms in the following. a) 1.30 mol of calcium hydroxide b) 0.75 mol of Au2(SO4)3

c) 2.25 mol of Cr(OH)3 d) 0.46 mol of SnF4

6. Find the mass of the following. a) 3.6 mol of Zn atoms b) 0.43 mol of He atoms c) 2 mol of CBr4

d) 3 mol of Cl2 e) 3.6 mol of Zn(OH)2 f) 0.75 mol of Pb3(AsO4)2

7. Determine the number of moles in the following. a) 35.00 g of Na b) 77.0 g of CuO

c) 88 g of Al2(SO4)3 d) 0.95 g of O2

8. Determine the mass of the following. a) 5.65 x 1024 molecules of CaBr2 b) 4.25 x 1023 molecules of N2

c) 9.10 x 1022 molecules of Ba(NO3)2 d) 3.01 x 1022 molecules of Ga2(CO3)3

9. Determine the number of molecules in each. a) 77.0 g of CuO b) 88.0 g of Al2(TeO4)3

c) 2.50 g of Ne d) 71.0 g of MgS

10. Determine the number of atoms in each. a) 85.0 g of Mn(OH)2 b) 125 g of CuSO4

c) 375 g of Sn(TeO4)2 d) 26.0 g of SO3

Complete the following Textbook questions.

11. P. 275 # 1 a, f, g

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SCH3  U-­‐  R.H.  KING  ACADEMY                                                        Moles  &  Stoichiometry  WORKSHEET   Name:        

W3    

Mole Relationships in Chemical Reactions

The quantitative relationship among reactants and products is called stoichiometry. The term stoichiometry is derived from two Greek words: stoicheion (meaning "element") and metron (meaning "measure"). Stoichiometry calculations are based on the fact that atoms are conserved. They cannot be destroyed or created. Numbers and kinds of atoms before and after the reactions are always the same. This is the Law of Conservation of mass and is why chemical reactions must be balanced. The molar coefficients from a balanced chemical equation represent the ratio in which reactants come together and make products in terms of moles.

For example, in the reaction 2 H2(g) + O2(g) 2 H2O(g) one mole of oxygen gas reacts with two moles of hydrogen gas to make 2 moles of water.

In most stoichiometry problems, amounts will be given in something other than moles and therefore you must always convert to moles before working with the molar ratio given by the balanced chemical equation. The following flow chart outlines the basic process to any stoichiometry calculation.

Steps to Solving a Stoichiometry Problem

1. Write a balanced chemical equation for the reaction. 2. Identify the known and unknown materials in the problem. 3. Calculate the number of moles of the known or given material. 4. Use the molar ratio from the balanced chemical equation to determine the number of moles of the

unknown. 5. Convert the number of moles of the unknown material to whatever is asked for in the question.

Sample Problem: Calculate the mass of oxygen gas produced by the decomposition of 12.26 g of potassium chlorate into potassium chloride and oxygen.

Solution: 2 KClO3 2 KCl + 3 O2 12.26 g ?

moles  KClO! =mass

MM  KClO!=  

12.26  g122.55  g/mol

= 0.1000  mol

moles  of  O!moles  of  KClO!

=  32=  

x0.1000  mol

moles  of  O! =3  (0.1000)

2=  0.1500  mol

mass  of  O! = nMM = 0.1500  mol 32.00gmol

mass  of  O! = 4.800  g

mass  of  material  

concentration  and  volume  of  material  in  

solution  

P  V  T  of  a  gaseous  material  

moles  of  known  material  

use  the  molar  ratio  in  the  balanced  chemical  equation  to  determine  the  number  of  moles  of  the  unknown  material  

moles  of  unknown  material  

mass  of  unknown  

concentration  and  volume  of  unknown  material  in  solution  

P  V  T  of  a  gaseous  unknown  material  

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SCH3  U-­‐  R.H.  KING  ACADEMY                                                        Moles  &  Stoichiometry  WORKSHEET   Name:        

W4    

Exercise

Complete a full solution for each problem and express your final answer to the correct number of significant digits.

1. Calcium hydroxide reacts with carbon dioxide to produce calcium carbonate and water. What mass of calcium carbonate would be produced from 246.4 g of calcium hydroxide? Remember to write the balanced chemical equation first. [332.8 g]

2. What mass of carbon dioxide are formed when 48.0 g of carbon reacts with oxygen gas to produce

carbon dioxide? [176 g]

3. For the reaction 6 CO2(g) + 6 H2O(g) C6H12O6(s) + 6 O2(g), determine the following. A) What mass of water is needed to produce 50.0 g of glucose (C6H12O6)? [30.0 g] B) How many grams of carbon dioxide are needed to react with 27.0 g of water? [65.9 g]

4. For the reaction: Cu(s) + 2 AgNO3(aq) Cu(NO3)2(aq) + 2 Ag(s)

A) How many grams of copper can react with 10.0 g of silver nitrate? [1.87 g] B) What mass of silver is formed when 5.0 g of copper reacts with excess silver nitrate? [17 g] C) Calculate the mass of copper(II) nitrate produced by 1.0 g of silver nitrate? [0.55 g]

5. Zinc metal reacts with hydrochloric acid to produce zinc chloride and hydrogen gas. What mass of zinc

chloride is produced by the reaction of 2.3 g of zinc metal? [4.8 g]

6. Zinc reacts with sulfuric acid to form hydrogen gas and zinc sulfate. What mass of hydrogen gas is produced from 31.8 g of zinc? [0.982 g]

7. What mass of sulfurous acid can be produced when 128 g of sulfur dioxide combines with water? [164 g]

8. Silver bromide is formed when silver nitrate reacts in a double displacement reaction with sodium

bromide. What mass of silver bromide can be produced from 34.3 g of sodium bromide? Remember sodium nitrate is also formed. [62.6 g]

9. Hydrochloric acid is added to 50.0 g of iron (II) sulfide. What mass of hydrogen sulfide is produced? Remember iron (II) chloride will also be formed. [19.4 g]

10. How much nitric acid is needed to react completely with 25 g of magnesium to produce hydrogen gas

and magnesium nitrate? [130 g]

11. What mass of cupric chloride can be produced from 75.0 g of cupric oxide in the following reaction? CuO + 2 HCl CuCl2 + H2O [127 g]

12. When aluminium is heated in oxygen, aluminium oxide is formed. What mass of the oxide can be

formed from 25.0 g of the metal? [47.2 g]

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SCH3  U-­‐  R.H.  KING  ACADEMY                                                        Moles  &  Stoichiometry  WORKSHEET   Name:        

W5    

Factor-label Method: 1. How many molecules of bromine are present in 0.045 mole of bromine gas? [2.7 x 10 22 of Br2

molecules]

2. How many moles of carbon atoms are present in 1.85 moles of glucose? [22.2 moles of C atoms]

3. Ethylene glycol (antifreeze) has the formula C2H6O2. How many molecules are present in a

3.86 × 10-20 g sample? [374 molecules]

4. A chemical reaction requires 3.000 moles of sodium chloride. How many grams is this? [175.3g]

5. How many moles of H2 are in 100 g of H2? [49.5 mol]

6. 300 g of CuSO4 is needed in an experiment. How many moles does this represent? [1.88 mol]

7. A chemical reaction requires 23.78 moles of silver chloride. How many grams is this? [3408 g]

8. How many molecules are in 73 grams H2O? (hint: form a conversion factor using Avogadro’s #) [2.4 x 10 24 of H2O molecules]

9. 255 g of calcium phosphate are produced in a chemical reaction. How many moles of calcium

phosphate does this represent? How many moles of phosphorus are produced? [0.822 mol of calcium phosphate, 1.64mol of phosphorus]

10. It's starting to get cold, and everybody is going to want to put on their boots. The equation for the

reaction is as follows: 1(miserable cold person) +2(warm boots) —> 1(happy warm person with boots)

Let's assume the average miserable cold person has a mass of 55 kg, and one warm boot has a mass of 0.5 kg.

A. If you had 330 kg of miserable cold people, how many miserable cold people would you have? How

many boots would you require to turn 330 kg of miserable cold people into happy warm people with boots? What mass of boots is this? [6.0 misearable cold people ; 6kg of boots; 12 boots]

B. What mass of happy warm people could you make with 3 kg of boots? Assume you have enough

miserable cold people to use up all of the boots. [3 happy warm people]

C. If you had 605 kg of miserable cold people, and 10 kg of boots, what mass of happy warm people could you make, assuming nobody is wearing only one boot? [560kg of hwp]

Mixed Mole Practice Problems:

12. Prozac, C17H18F3NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. Find its molar mass. [309.36 g/mol]

13. How many grams are in 2.50 moles of S? [80 g]

Page 6: Avogadro’s Number, Moles and Molar Masssch3uking.wikispaces.com/file/view/MOLES+&+STOICHIOMETRY+WO… · Avogadro’s Number, Moles and Molar Mass For a scientist to carry out experiments

SCH3  U-­‐  R.H.  KING  ACADEMY                                                        Moles  &  Stoichiometry  WORKSHEET   Name:        

W6    

14. How many atoms are present in 11 grams of Si? [2.4 x 10 23 of Si atoms]

15. The artificial sweetener aspartame (Nutra-Sweet) formula C14H18N2O5 is used to sweeten diet foods,

coffee and soft drinks. How many atoms of hydrogen of aspartame are present in 225 g of aspartame? [8.28 x 10 24 of H atoms]

16. What is the mass in grams of 1 water molecule? [2.99x 10 -23 g]

17. How many moles of atoms are in 100.0 grams of nitrogen gas, N2?[ 7.138 atoms]

18. How many ions are there in 28.7 grams of lithium nitrate, LiNO3?[ 5.01 x 10 23 ions]

19. How many moles of cations are there in 1.75 kg of magnesium chloride, MgCl2? [18.4 mol of cations]

20. How much would 3.80 x10 25 moecules of oxygen gas weigh? [2.02 x 10 3 g]

21. How many electrons are there in 75.0 grams of pure iron, Fe? [2.10 x 10 25 electrons]

22. What is the formula mass of cupric sulfate petahydrate, CuSO4 · 5H2O? 249.72g/mol] Complete the following Textbook questions.

23. P. 277 # 1, 7; P. 280 # 4; 282 # 2, P. 283 # 8

Fun with Moles

1. How many moles of atoms are found in the following? Show all calculations. a) 2 moles of H2O molecules b) 0.33 mol of C6H12O6 c) 35.0 g of Rb

d) 88.0 g of CO2 e) 148 g of Ca(OH)2 f) 48.0 g of oxygen gas

2. Calculate the mass in grams of the following.

a) 0.50 mol of oxygen gas b) 2.5 mol of elemental sulfur c) 0.75 mol of aluminum silicate

d) 1.30 mol of gold(III) chromate e) 1 water molecule f) 1 atom of carbon

3. Determine the number of particles (molecules for molecular compounds, formula units for ionic

compounds and atoms for elements). a) 25.0 g of hydrogen gas b) 34.0 g of NH3 c) 250 g of barium

d) 675 g g of nickel (III) dichromate e) 0.85 g of potassium f) 125 g of SnI4

4. Determine the number of moles of:

a) oxygen atoms in 64 g of ozone gas, O3 b) hydrogen atoms in 0.33 mol of NH3 c) hydrogen atoms in 45 g of water

d) electrons in 54 g of aluminum e) valence electrons in 85 g of sodium. f) ions in 175 g of MgCl2

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SCH3  U-­‐  R.H.  KING  ACADEMY                                                        Moles  &  Stoichiometry  WORKSHEET   Name:          

W7    

Percentage Composition

Percentage composition is the percent, by mass, of each element in a compound. This value gives the fraction of each element in a compound and can be calculated 2 ways.

1. Using Experimental Data:

With this method, the total mass of a sample of a compound must be known as well as the mass of each element. To calculate the percent composition of an element, use the formula below.

%  composition  of  an  element =mass  of  elementmass  of  compound

 x  100%

Example:

A 10.5 g sample of glucose was combusted to yield 4.20 grams of carbon, 5.60 grams oxygen and 0.70 grams of hydrogen. Calculate the percent composition for each element in the glucose sample.

Solution:

%  C = 40.0%

%  O = 53.3%

%  H = 6.7%

2. Using the Chemical Formula of a Compound:

With this method the molar mass of the compound must be calculated and the total mass of each element (# of atoms on an element multiplied by the atomic mass of the element). To calculate the percent composition of an element, use the formula below.

Example:

Determine the percent composition of each element in phosphoric acid (H3PO4).

Solution:

MM of H3PO4= 3 H + P + 4 O MM of H3PO4= 3(1.01 g/mol) + 30.97 g/mol + 4(16.00 g/mol) MM of H3PO4=98.00 g/mol

%  H = 3.09%

%  P = 31.60%

%  O = 65.31%

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SCH3  U-­‐  R.H.  KING  ACADEMY                                                        Moles  &  Stoichiometry  WORKSHEET   Name:      

W8    

Exercise: Complete the following on a separate piece of paper and express the answer to the correct number of significant digits.

1. When 2.00 g of lead is heated with sulfur, 2.30 g of lead (II) sulfide is formed. Calculate the percent composition by mass of each element in lead (II) sulfide. [87.0 % Pb, 13.0 % S]

2. Mercury (II) oxide decomposes to its elements when heated. Use the data below to calculate the percent composition of each element. [92.30 % Hg, 7.69 % O]

mass of test tube 12.25 g mass of test tube and mercury (II) oxide 13.68 g mass of test tube and mercury 13.57 g

3. Iron and sulfur react to form iron (II) sulfide. Use the data below to calculate the percent composition of each element. [63.6% Fe, 36.4% S]

mass of crucible 30.6 g mass of crucible and iron 41.8 g mass of crucible and iron (II) sulfide 48.2 g

4. A compound contains 30.1 % oxygen by mass. What mass of oxygen is there in 5.0 g of the compound? [1.51 g]

5. Epsom salt is the hydrated form of magnesium sulfate and can be dehydrated by heating. Use the following data to calculate the percent of water in Epsom salt. [51.1 %]

mass of crucible 40.88 g mass of crucible and Epsom salts 65.52 g mass of crucible and dry magnesium sulfate 52.92 g

6. Calculate the percent composition by mass of all elements in each compound a) H2O2 [5.91% H, 94.09 % O] b) CO [42.89% C, 57.11 % O] c) Na2SO4

d) HNO3 e) Pb(NO3)2 f) C6H12O6

7. An unknown compound has a molar mass of 18.02 g/mol. It is known to contain 88.9% oxygen and

11.1 % hydrogen. What is the compound? [H2O]

8. An unknown compound contains 7.17 % oxygen and 92.83% Pb. If the molar mass of the compound is 223.2 g/mol, what is the formula and name of the compound? [PbO, lead (II) oxide]

9. An unknown compound contains 69.55 % oxygen and 30.45 % nitrogen. If the molar mass of the compound is 92.02 g/mol, what is the formula and name of the compound? [N2O4]

10. P. 288 # 4, 6

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SCH3  U-­‐  R.H.  KING  ACADEMY                                                        Moles  &  Stoichiometry  WORKSHEET   Name:        

W9    

Empirical Formula Calculations

1. Methane gas was determined to be 75 % C and 25 %H by mass. Determine its empirical formula.

2. Calculate the simplest formula for the compounds whose compositions are listed below. a) 15.75 % C, 84.25 % S

b) 31.13 % S, 68.87 % Cl

c) 81.61 % Cl, 18.39 % O

d) 70.11 % Ag, 9.10 % N, 20.79 % O

e) 40.27 % K, 26.78 % Cr, 32.95 % O

f) 26.58 % K, 35.35 % Cr, 38.07 % O

3. An analysis of nicotine gave 74.07% C, 8.64 % H, 17.28 % N. What is the simplest formula for nicotine?

4. A hydrated compound was found to contain 13.92 % Ca, 1.39 % H, 21.52 % P, 44.43 % O and 18.74 % water. What is the simplest formula of the compound?

5. A 7.81 g sample of an unknown compound was broken down to give 4.60 g of sodium and 3.21 g

of sulfur. What is the simplest formula for this compound?

6. When the oxides of mercury are heated, they decompose to mercury and oxygen gas. If a 1.083 g sample of an oxide of mercury produces 1.003 g of mercury, what is the empirical formula of this compound?

7. A student wishes to find the formula of a compound formed when copper is heated is heated in the

presence of molten sulfur. Use the data below collected by the student to determine the simplest formula.

mass of empty crucible and lid 24.60 g mass of crucible, lid and copper 25.87 g mass of crucible, lid and product 26.19 g

8. It was determined that a 10.66 g sample of a pure substance contained 3.91 g of potassium, 3.55 g of chlorine and 3.20 g of oxygen. What is the empirical formula of this compound?

9. The element “M” forms the chloride MCl4. If the chloride contains 75 % chlorine by mass, calculate the molar mass of element “M”.

10. The elements “X” and “Y” combine to form a compound which is 40 % “X” by mass. If the molar mass of “X” is twice that of “Y”, what is the simplest formula of the compound formed?

Answers:

1. CH4 2. a) CS2

b) SCl2 c) Cl2O d) AgNO2 e) K2CrO4

f) K2Cr2O7 3. C5H7N 4. CaH4P2O8 · 3 H2O or

Ca(H2PO4)2 · 3 H2O 5. Na2S 6. HgO

7. Cu2S 8. KClO2 9. 47.3 g/mol 10. XY3

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Molecular or True Formula Calculations

The molecular or true formula of a compound is a multiple of the empirical formula. To determine the molecular or true formula, it is necessary to know both the empirical formula for the compound and the molar mass of the compound. The steps to determine the molecular or true formula are:

1. Determine the empirical (simplest) formula. 2. Calculate the molar mass of the compound (if not given). 3. Calculate the molar mass of the empirical formula. 4. Find the number of empirical formulas in the true formula (MM true formula ÷ MM empirical

formula) 5. Multiply the empirical formula by the number of empirical formulas.

Example:

A compound with a molar mass of 42.5 g/mol has the simplest formula CH2. What is the molecular or true formula?

Solution: molar mass of empirical formula = 12.01 g/mol + 2(1.01 g/mol) = 14.03 g/mol

#  of  empirical  formulas =MM  of  true  formula

MM  of  empirical  formula=

42.5 gmol

14.03 gmol

 

#  of  empirical  formulas = 3.03

Therefore the true formula is C3H6.

Exercise: 1. The simplest formula and molar masses for several compounds are listed below. Calculate the true

formula in each case. a) CH2 MM = 85 g/mol b) C5H6 MM = 197 g/mol

c) NO2 MM = 92 g/mol d) C4H4NO2 MM = 96 g/mol

2. The percent composition by mass and approximate molar mass of some compounds are listed below. Calculate the molecular formula in each case. a) 52.2 % C, 13.0 % H, 34.8 % O MM = 90 g/mol b) 40.3 % B, 52.2 % N, 7.5 % H MM = 80 g/mol

3. Two substances, one gas and the other a volatile liquid have the same percent composition by mass;

92.3 % C, 7.7 % H. The molar mass of the gas is approximately 25 g/mol and that of the liquid is 79 g/mol. Determine the true formula of each compound.

4. A compound was analyzed and found to be 40.00 % C, 6.67 % H, 53.33 % O. It was also found that

3.01 x 1021 molecules of this substance had a mass of 0.60 g. Find the true formula of the compound. Answers:

1. a) C6H12 b) C15H16

c) N2O4 d) C4H4NO2

2. a)C4H12O2 b) B3N3H6

3. C2H2 and C6H6 4. C4H8O4

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Empirical and True Formulas Extra Practice

1. Determine the empirical formula for each of the following. a) 8.04 % Li, 91.96 % Br b) 70.9 % K, 29.1 % S

c) 28.6 % Mg, 14.3 % C, 57.1 % O d) 29.4 % Ca, 23.5 % S, 47.1 % O

2. Find the empirical formulae of the following compounds:

a) A compound containing 4.0 g of hydrogen and 32.0 g of oxygen b) A compound containing 24.0 g of calcium and 5.6 g of nitrogen

3. A 16.4 g sample of hydrated calcium sulfate is heated until all the water is driven off. The calcium

sulfate that remains has a mass of 13.0 g. Find the formula and chemical name of the hydrate.

4. When 8.00 g of Pb(C2H3O2)2 · x H2O are heated, 1.14 g of H2O are driven off. Find the chemical formula and the name of the hydrate.

5. A hydrate is determined to be 45.43% water and 54.57% CoCl2. Find the chemical formula and name

for this hydrate.

6. Find the empirical formulas of the compounds formed when: a) 4.00 g of mercury forms 4.64 g of a mercury sulphide b) 0.62 g of phosphorus forms 4.17 g of a phosphorus chloride

7. Find the empirical and molecular formulas for each of the following compounds.

a) 51.28 % C, 9.40 % H, 27.35 % O, 11.97 % N MM = 234 g/mol b) 33.18 % C, 4.60 % H, 29,49 % O, 32.72 % Cl MM = 108.5 g/mol c) 45.71 % C, 10.48 % H, 30.48 % O, 13.3 % N MM = 210 g/mol

8. The empirical formula for trichloroisocyanuric acid, the active ingredient in many household bleaches, is

OCNCl. The molar mass of this compound is 232.41 g/mol. What is the molecular formula of trichloroisocyanuric acid?

9. Determine the true formula of the compound with an empirical formula of CH and molar mass of 78.11 g/mol.

10. What is the true formula of a compound with a molar mass of 108 g/mol, made from 4.04 g of N and 11.46 g O?

11. P. 293 # 10; P.300 # 4; P. 310 # 57,72,82

12. Answers: 1. a) LiBr

b) K2S c) MgCO3 d) CaSO4

2. a) H2O b) Ca3N2

3. CaSO4 · 2 H2O 4. Pb(C2H3O2)2 · 3 H2O 5. CoCl2 · 6 H2O 6. a) HgS

b) PCl5

7. a) C5H11O2N, C10H22O4N2 b) C3H5O2Cl, C3H5O2Cl c) C4H11O2N, C8H22O4N2

8. O3C3N3Cl3 9. C6H6 10. N2O5

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Stoichiometry: Follow the rules for significant digits. Show all calculations.  

1. 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O a) What mass of O2 will react with 400.0 g C4H10? [1431 g] b) How many moles of water are formed in a)? [34.40 mol]

2. 3 HCl + Al(OH)3 -> 3 H2O + AlCl3

How many grams of aluminum hydroxide will react with 5.30 moles of HCl? [138 g] 3. Ca(ClO3)2 -> CaCl2 + 3 O2

What mass of O2 results from the decomposition of 1.00 kg of calcium chlorate? [464 g] 4. The reaction of Ca with water can be predicted using the activity series. What mass of water is

needed to completely react with 2.35 g of Ca? [1.14g] 5. Fe2O3 + 3CO -> 2Fe + 3CO2. a) How many moles of carbon monoxide are required to react with 163.0 g of iron (III) oxide? [3.062 mol CO] b) How many grams of CO2 are produced from a reaction that also produces 23.9 grams of Fe? [28.2 g CO2] 6. 3Cu + 8HNO3 → 3Cu(NO3)2 + 4H2O + 2NO a) How many moles of copper (II) nitrate can be prepared from 17.0 moles of Cu? [17.0 mol] b) How many grams of copper (II) nitrate can be prepared using 3.8 moles of HNO3? [230 g] c) What mass of water results from the reaction of 8.50 kg of copper metal? [3.12 kg] 7. According to the following equation how many moles of lithium hydroxide are required to react with

20.0 moles of CO2? CO2 + 2LiOH → Li2CO3 + H2O [40.0 mol] 8. How many moles of ammonia, NH3 are produced when 6.00 moles of hydrogen gas react with excess

nitrogen gas? [4.00 mol] 3H2 + N2 → 2NH3

9. What mass, in grams, of glucose is produced when 3.00 moles of water react with carbon dioxide?

6CO2 + 6H2O → C6H12O6 + 6O2 [90.1 g] 10. What mass in grams of magnesium oxide is produced when 2.00 moles of magnesium react with

oxygen? [80.6 g] 11. The reaction below is run using 824.0 grams of NH3 and excess oxygen, how many moles of NO are

formed? How many moles of H2O are formed? [48.38 g NO; 72.58 mol H2O] 4NH3 + 5O2 → 4NO + 6H2O 12. If mercury (II) oxide decomposes, how many grams of mercury (II) oxide are needed to produce 125 grams of oxygen? [1690g]

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13. How many grams of SnF2 are produced from the reaction of 30.0 grams of HF with Sn? [118 g] Sn + 2HF → SnF2 + H2

14. What mass of aluminum is produced by the decomposition of 5000.0 grams of Al2O3? [2.646 kg] 15. How many molecules of hydrogen chloride gas will be produced when 33.5 grams of chlorine combine

with excess hydrogen? [5.69 x1023 molecules]

Complete the following Textbook questions

16.      P.  325  #  2,5,  8  (b)    

Stoichiometry- Limiting Reagents Follow the rules for significant digits. Show all calculations.  

1. 2Al + 6HCl → 2AlCl3 + 3H2 : If 25.0 g aluminum was added to 90.0 g HCl, what mass of H2 will be produced? [2.49 g H2]

 

2. N2 + 3H2 → 2NH3 : If you have 20.0 g of N2 and 5.0 g of H2, which is the limiting reagent? [N2]  

3. 4Al + 3O2 → 2 Al2O3 : What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O2? [18.9 g Al2O3]

 

4. C3H8 + 5O2 → 3CO2 + 4H2O: When C3H8 burns in oxygen, CO2 and H2O are produced. If 15.0 g of C3H8 reacts with 60.0 g of O2, how much CO2 is produced? [44.9 g CO2]

 

5. MgCl2 + 2AgNO3 → Mg(NO3)2 + 2AgCl: If 25 g magnesium chloride was added to 68.0 g silver nitrate, what mass of AgCl will be produced? [57 g AgCl]

 

6. Silicon dioxide (quartz) is usually quite uncreative but reacts readily with hydrogen fluoride according to the following equation. SiO2 (s) + 4HF (g) → SiF4 (g) + 2H2O (l): If 2 moles of HF are exposed to 4.50 moles of SiO2, which is the limiting reactant? [HF]

 

7. If 20.50 grams of chlorine is reacted with 20.50 grams of sodium, which reactant is in excess? [Na]  

8. Using what you learned from problem #7, how much sodium chloride would be produced in 20.5 grams of chlorine is reacted with 20.50 grams of sodium? [33.8 g NaCl]

 

9. Magnesium sulfate reacts with potassium. How much magnesium would be produced if 35.0 grams of magnesium sulfate is exposed to 45.0 grams of potassium? [7.07 g Mg]

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W14    

10. What is the maximum number of grams of PH3 that can be formed when 6.20g of solid phosphorus reacts with 6.0g of hydrogen gas to form PH3 gas? [6.8 g PH3]

 

11. Solid copper is formed when solid aluminum reacts with aqueous copper (II) sulfate in a single-replacement reaction. How many grams of copper can be obtained when 29.0g of Al react with 156.0g of copper (II) sulfate? [62.1g Cu]

 

12. If you begin with 1250g of nitrogen gas and 225.0g of hydrogen gas in the reaction that forms ammonia gas (NH3), how much ammonia will be formed? What is the limiting reagent? How much of the reagent is left when the maximum amount of ammonia is formed? [1265 g NH3, 210 g N2]

13. 550.0g of aqueous lithium hydroxide reacts with 550.g of aqueous aluminum nitrate in a double replacement reaction. How many grams of solid aluminum hydroxide are produced? What is the limiting reactant? [ Al(NO3)3 – the limiting reactant, 201 g of Al(OH)3]

Complete the following Textbook questions

14.      P.  330    #  2,  4(a),  6;  P.335  #1,  4,    

Percent Yield: 1. The electrolysis of water forms H2 and O2. What is the % yield of O2 if 12.3 g of O2 is produced from

the decomposition of 14.0 g H2O? [12.43 g; 98.9%] 2. 107 g of oxygen is produced by heating 300 grams of potassium chlorate. Calculate % yield. [117.5

g; 91.1%]

3. What is the % yield of ferrous sulphide if 3.00 moles of Fe reacts with excess sulfur to produce 220 grams of ferrous sulphide? Fe + S → FeS [263.7 g; 83.4%]

4. Iron pyrites (FeS2) reacts with oxygen according to the following equation: 4FeS2 + 11O2 → 2Fe2O3 + 8SO2

If 300 g of iron pyrites is burned in 200 g of O2, 143 grams of ferric oxide is produced. What is the percent yield of ferric oxide? [181.48 g Fe2O3; 78.8%]

5. 70 grams of manganese dioxide is mixed with 3.5 moles of hydrochloric acid. How many grams of Cl2

will be produced from this reaction if the % yield for the process is 42%? [57.08 g Cl2; (42% x 57.08 g Cl2)/(100%) = 24 g Cl2]

MnO2 + 4HCI → MnCl2 + 2H2O + Cl2

Complete the following Textbook questions

6.       P.  339      #  4,  5,  13;  

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REVIEW

For all questions calculate molar masses to two decimal places and give answers with the correct number of significant digits (remember: do not round your values until writing the final answer).

MOLES

1. Give the percentage composition for each compound, by mass: a) H2SO4, b) Ca(OH)2.

2. Calculate the molar mass of a) H2SO4, b) Fe2(Cr2O7)3.

3. a) How many moles are in 16 grams of CuCl2? b) How much does 70 moles of NaCl weigh?

4. a) How many molecules are in exactly 4 moles of H2O? b) How many hydrogen atoms are in exactly 4 moles of H2O? c) How many hydrogen atoms are in 0.173 moles of H2O?

5. What mass of magnesium oxide results when 56.3 g O2 combines with excess magnesium?

6. Label as simplest formula, molecular formula, or both: a) CuCl2, b) CO2, c) O2, d) C4H10.

7. A substance is 80% C and 20% hydrogen by mass. a) What is the simplest formula? b) What is the molecular formula of the compound if the molar mass is 30 g/mol?

8. Balance these equations: a) C40H82 + O2 → CO2 + H2O, b) H2O + Al4C3 → CH4 + Al(OH)3

9. What four things may cause actual yields to differ from theoretical yields?

STOICHIOMETRY

10. Sodium and water react. How many grams of sodium will react with water to produce 4.00 moles of hydrogen? (184 grams)

11. How many moles of lithium chloride will be formed by the reaction of chlorine with 4.00 grams of lithium bromide? (0.0460 mol or 4.60X10-2 mol)

12. Aluminum will react with sulfuric acid. a. How many moles of H2SO4 will react with 18 mol of aluminum? b. How many moles of each product will form? (a. 27 mol; b. 27 mol hydrogen and 9.0 mol aluminum sulfate)

13. What mass of acetylene, C2H2, will be produced from the reaction of 90. grams of calcium carbide with water in the following reaction? (37 grams)

CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(s)

14. What mass of ZrCl4 can be produced if 862 grams of ZrSiO4 and 950. grams of chlorine are available to react according to the following equation? (1.10X103 grams)

ZrSiO4 + 2Cl2 → ZrCl4 + SiO2 + O2

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15. Heating zinc (II) sulfide in the presence of oxygen yields zinc (II) oxide and sulfur dioxide. If 1.72 moles of ZnS is heated in the presence of 3.04 moles of O2, which reactant will be used up? (ZnS is the limiting reactant)

16. Aluminum and oxygen react. a. Which reactant is limiting if 0.32 moles of aluminum and 0.26 moles of oxygen are available? b. How many moles of aluminum oxide are formed from the reaction of 0.00638 moles of oxygen and 0.00915 moles of aluminum? c. If 3.17 grams of aluminum and 2.55 grams of oxygen are available, which reactant is limiting? ( a. aluminum is limiting; b. 4.25 X 10-3 mol; c. oxygen is limiting)

17. Bromine replaces iodine in magnesium iodide during a single replacement reaction. a. Which is the excess reactant when 560 grams of MgI2 and 360 grams of Br2 react, and what mass remains? b. What mass of I2 is formed in the same process? (a. Br2 is the excess and 38 grams will remain; b. 510 grams)

18. If 8.87 grams of As2O3 is used in the below reaction and 5.33 gram of arsenic is produced, what is the present yield? (79.3%)

2As2O3 + 3C → 3CO2 + 4As

19. Tungsten can be produced from its oxide by reacting the oxide with hydrogen at a high temperature according to the following equation:

WO3 + 3H2 → W + 3H2O

a. What is the percent yield if 56.9 grams of WO3 yields 41.4 grams of tungsten? b. How many moles of tungsten will be produced from 3.72 grams of WO3 if the yield is 92.0%? (a. 91.8%; b. 0.0148 mol).

REVIEW - MOLES & STOICHIOMETRY

P. 354-362- read over all the questions, attempt as many as possible (write solutions to questions you

find challenging).