Chapter 3 Calculations Related to Chemical Formulas and ...profkatz.com/courses/wp-content/uploads/2017/01/CH1710-Lecture-7... · Chapter 3 Calculations Related to Chemical Formulas

Chapter 3

Calculations Related to Chemical Formulas and

Equations

Percent Composition

Percent Composition

Percentage of each element in a compound by mass

Can be determined from

1. the formula of the compound 2. the experimental mass analysis of the compound

1. Find the mass percent of Cl in C2Cl4F2

Mass Percent as a Conversion Factor2. If NaCl is 39% sodium, find the mass of table salt containing 2.4 g of Na.

100 g NaCl39 g Na

39 g Na100 g NaCl

2.4 g Na x = 6.1538 g NaCl 6.2 g NaCl 100 g NaCl39 g Na

g Na g NaCl

Empirical Formulas

Empirical Formula

Simplest, whole-number ratio of the atoms of elements in a compound

Can be determined from elemental analysis

Finding an Empirical Formula from % Composition

1. Convert the percentages to grams

2. Convert grams to moles

3. Write a pseudoformula using moles as subscripts

4. Divide all by smallest number of moles

5. Multiply all mole ratios by number to make all whole numbers

Example:Find the empirical formula of aspirin with the given mass percent composition

Given: C = 60.00%H = 4.48% O = 35.53%

Therefore, in 100 g of aspirin there are 60.00 g C, 4.48 g H, and 35.53 g O.

However, ratios of mass are not useful in determining empirical formulas !!

We must convert to a ratio of moles !!

g C g H g O

mol C mol H mol O

pseudoformulaCxHyOz

empirical formulaCxHyOz

Manipulate subscripts to obtain whole-number ratio

Methodology for determining empirical formula:

Calculate the moles of each element

Write a pseudoformula

C4.996H4.44O2.220

Find the mole ratio

C2.25H2.00O1.00Multiply subscripts by factor to give whole number

C9H8O4

(x 4)

÷ 2.220

C4.996H4.44O2.220

3. Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70 g/mol)

and the rest fluorine (19.00 g/mol)

Given: 75.7% Sn, (100 – 75.3) = 24.3% F

g Sn mol Sn

g F mol F

pseudo formula

empirical formula

Element Ratio in Grams Molar Mass Ratio in Moles Ratio in Moles

Sn 75.7g1mol

118.7g0.6377 1

F 24.3g1mol

19.00g1.279 2.005

6. Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70 g/mol)

and the rest fluorine (19.00 g/mol)

SnF2

X

X

÷0.6377

4. Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00)

Given: 72.4% Fe, (100 – 72.4) = 27.6% O

g Fe mol Fe

g O mol O

pseudo formula

empirical formula

7. Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00)

Element Ratio in Grams Molar Mass Ratio in Moles Ratio in Moles Ratio in Moles

Fe 72.4g1mol

55.85g1.296 1 3

O 27.6g1mol

16.00g1.725 1.33 4

Fe3O4

X

X

÷1.296 x 3

Chapter 3 Molecular Formulas

Molecular Formulas

The molecular formula is a multiple of the empirical formula.

To determine the molecular formula you need to know the empirical formula and the molar mass of the compound.

Find the molecular formula of butanedione if its empirical formula is C2H3O and its molar mass (MM) is 86.03 g/mol.

Factor of 2Molar Mass (emp. form.) = 2 x (12.01 gC/molC) + 3 x (1.008 gH/molH) + 1 x (16.00 gO/molO) = 43.04 g/mol

Molecular formula = C2H3O x 2 = C4H6O2

Practice – Benzopyrene has a molar mass of 252 g and an empirical formula of C5H3. What is its molecular formula?

(C = 12.01, H=1.01)

Molecular formula = {C5H3} x 4 = C20H12

C5 = 5(12.01 g) = 60.05 gH3 = 3(1.01 g) = 3.03 gC5H3 = 63.08 g

252

?

Combustion Analysis

Combustion Analysis

A known mass of compound is burned in oxygen and the masses of the products formed (CO2 and H2O) are determined.

By knowing the masses of the products and composition of constituent elements in the product, the original amount of

constituent elements can be determined.

It is assumed that all of the carbon in the original sample is converted to carbon dioxide and all of the

hydrogen in the sample is converted to water.

(Generally used for organic compounds containing C, H, O)

Combustion Analysis

Example of Combustion AnalysisCombustion of a 0.8233 g sample of a compound containing only

carbon, hydrogen, and oxygen produced the following:

CO2 = 2.445 g H2O = 0.6003 g

Determine the empirical formula of the compound.

This came from C.This came from H.

gCO2, H2O

molCO2, H2O

molC, H

gC, H

gO

molO

gC, H

pseudoformula empirical formula

molC,H,O

1 mole H = 1.008 g H 1 mole C = 12.01 g C molar masses of elements1 mole O = 16.00 g O

1 mole CO2 = 44.01 g CO2 ⇒1 mole H2O = 18.02 g H2O

1 mole C ⇒ 1 mole CO2 2 mole H ⇒ 1 mole H2O

In the original sample

molar masses of compounds

ratios of compounds formed in combustion




1 mole H = 1.008 g H 1 mole C = 12.01 g C molar masses of elements1 mole O = 16.00 g O


C0.05556H0.06662O0.00556

Pseudo formula

÷ 0.00556

Empiricalformula

1.92 g CO2 x x = 0.0436 mol C1 mol CO2 44.01 g CO2

1 mol C 1 mol CO2

0.784 g H2O x x = 0.0870 mol H1 mol H2O 18.02 g H2O

2 mol H 1 mol H2O

0.0870 mol H x = 0.0877 g H1.008 g H 1 mol H

0.0436 mol C x = 0.524 g C12.01 g C 1 mol C


5. Combustion of 0.844 g of caproic acid produced 0.784 g of H2O and 1.92 g of CO2. If the molar mass of caproic acid is 116.2 g/mol,what is the molecular formula of caproic acid?


0.844 g compound - (0.524 g C + 0.0877 g H) = 0.232 g O

0.232 g O x = 0.0.0145 mol O1 mol O 16.00 g O

C H O g 0.524 0.0877 0.232moles 0.0436 0.0870 0.0145


C0.0436H0.0870O0.0145

C3H6O1

0.0145 0.0145 0.0145

molecular weight based on empirical

formula

Actual molecular formula = {C3H6O} x 2 = C6H12O2

StoichiometryThe study of the numerical relationship between

chemical quantities in a chemical reaction

Making Pizza

The number of pizzas you can make depends on the amount of the ingredients you use.

1 crust + 5 oz. tomato sauce + 2 cu cheese ➜ 1 pizza

This relationship can be expressed mathematically 1 crust : 5 oz. sauce : 2 cu cheese : 1 pizza

If you want to make more than one pizza, you can use the amount of cheese you have to determine the number of pizzas you can make.

Predicting Amounts from Stoichiometry

The amounts of any other substance, which can theoretically produced or consumed in a chemical

reaction, can be determined from the amount of just one substance.

Moles of A

Moles of B

Grams of A

Grams of B

Particles of A

Particles of B

Avogadro’s Number Avogadro’s Number

Molar MassMolar Mass

Coefficients

Grams of B

Grams of A

Moles of A

Moles of B

Mole to Mole Ratio from balanced equation

Stoichiometry Road Map

Molar Mass

Particles of B

Avogadro’s Number

Particles of A

1 mol glucose 6 mol water

6 mol water 1 mol glucose conversion factors

glucose + oxygen gas ➜ carbon dioxide + water

C6H12O6 + 6 O2 ➜ 6 CO2 + 6 H2O

1 mol C6H12O6 6 mol H2O

6 mol H2O 1 mol C6H12O6 conversion factors

mol H2O mol C

6H

12O

6

1. According to the following equation, how many moles of water are made in the combustion of 0.10 moles of glucose?

0.10 mol C6H12O6 x = mol H2O6 mol H2O1 mol C6H12O6 0.60

2. Estimate the mass of CO2 produced in 2007 by the combustion of 3.5 x 1015 g of octane (C8H18).

C8H18(l) + O2(g) ➜ CO2(g) + H2O(g)

2 C8H18(l) + 25 O2(g) ➜ 16 CO2(g) + 18 H2O(g)

g C8H18 mol CO2 g CO2 mol C8H18

2 mol C8H18 16 mol CO2

16 mol CO2 2 mol C8H18 conversion factors

1 mol C8H18 114.22 g C8H18

114.22 g C8H18 1 mol C8H18 conversion factors

1 mol CO2 44.01 g CO2

44.01 g CO2 1 mol CO2 conversion factors

3.5 x 1015 g C8H18 x x x

= 1.0789 x 1016 g CO2

16 mol CO22 mol C8H18

44.01 g CO21 mol CO2

1 mol C8H18114.22 g C8H18

2. Estimate the mass of CO2 produced in 2007 by the combustion of 3.5 x 1015 g of octane (C8H18).

2 C8H18(l) + 25 O2(g) ➜ 16 CO2(g) + 18 H2O(g)

2 mol C8H18 16 mol CO2

16 mol CO2 2 mol C8H18 conversion factors

1 mol C8H18 114.22 g C8H18

114.22 g C8H18 1 mol C8H18 conversion factors

1 mol CO2 44.01 g CO2


1.1 x 1016

g C8H18 mol C8H18 mol CO2 g CO2

1 mol C6H12O6 6 mol CO2

6 mol CO2 1 mol C6H12O6

conversion factors

3. How many grams of glucose can be synthesized from 37.8 g of CO2 in photosynthesis?

6 CO2 + 6 H2O ➜ C6H12O6 + 6 O2

37.8 g CO2 x x x

= 25.796 g C6H12O6

1 mol CO244.01 g CO2

1 mol C6H12O66 mol CO2

180.2 g C6H12O61 mol C6H12O6

1 mol CO2 44.01 g CO2


1 mol C6H12O6 180.2 g C6H12O6

180.2 g C6H12O6 1 mol C6H12O6 conversion factors

25.8

g CO2 mol CO2 mol C6H12O6 g C6H12O6

32.00 g O21 mol O2

1 mol O22 mol PbO2

1 mol PbO2239.2 g PbO2

100.0 g PbO2 x x x

= 6.68896 g O2

4. Lead (IV) oxide decomposes to yield lead(II) oxide and oxygen gas. How many grams of O2 can be made from the

decomposition of 100.0 g of PbO2?

g O2 mol PbO2 g PbO2 mol O2

2 PbO2(s) → 2 PbO(s) + O2(g)

(PbO2 = 239.2, O2 = 32.00)

1 mol PbO2 239.2 g PbO2

1 mol O2 2 mol PbO2

32.00 g O2 1 mol O2

6.689

The Limiting Reagent

More Making Pizzas1 crust + 5 oz. tomato sauce + 2 cu cheese ➜1 pizza

What would happen if we had 4 crusts, 15 oz. tomato sauce, and 10 cu cheese?

Limiting reagent Theoretical

yield

Limiting and Excess Reactants in the Combustion of Methane

CH4(g) + O2(g) ➜ CO2(g) + H2O(g)

CH4(g) + 2 O2(g) ➜ CO2(g) + 2 H2O(g)

➜

➜

If we have five molecules of CH4 and eight molecules of O2, which is the limiting reactant?

8 mol O2 x = 4 mol of CO21 mol CO22 mol O2

The Limiting Reactant

For reactions with multiple reactants, it is likely that one of the reactants will be

completely used before the others.

When this reactant is used up, the reaction stops and no more product is made.

1.00 mol Si3N4 2.00 mol N2

1.00 mol N2 x = 0.500 mol Si3N4

1.20 mol Si x = 0.400 mol Si3N41.00 mol Si3N4 3.00 mol Si

5. How many moles of Si3N4 can be made from 1.20 moles of Si and 1.00 mole of N2 in the reaction:

3 Si + 2 N2 ➜ Si3N4 ?

Theoretical yield

Limitingreactant

Percent Yield

More Making Pizzas

Let’s now assume that as we are making pizzas, we burn a pizza, drop one on the floor, or other uncontrollable events happen so that we only make two pizzas. The actual amount of product made in a chemical

reaction is called the actual yield.

We can determine the efficiency of making pizzas by calculating the percentage of the maximum number of

pizzas we actually make. In chemical reactions, we call this the percent yield.

6. When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield,

and percent yield.

TiO2(s) + 2 C(s)➜Ti(s) + 2 CO(g)

smalleramount is

from limitingreactant}

gC

kg

g

molC

g

mol

molTi

mol

mol

kgC

kgTiO2

gTiO2

molTiO2

molTi

Theoretical Yield

smaller molTi

gTi

kgTi

Theoretical Yield

Actual Yield

= % Yield

When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and

percent yield.

TiO2(s) + 2 C(s)➜Ti(s) + 2 CO(g)

Collect needed relationships:

1000 g = 1 kg Molar Mass Ti = 47.87 g/mol Molar Mass C = 12.01 g/mol

Molar Mass TiO2 = 79.87 g/mol


percent yield.

TiO2(s) + 2 C(s)➜ Ti(s) + 2 CO(g)

1 mole TiO2 : 1 mol Ti 2 mole C : 1 mol Ti

88.2 kg TiO2 x x x

= 1.1043 x 103 mol Ti

limiting reactantsmaller # moles of Ti Theoretical yield

When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield.

TiO2(s) + 2 C(s)➜Ti(s) + 2 CO(g)

1.10 x 103

28.6 kg C x x x

= 1.1907 x 103 mol Ti

1000 g C1 kg C

1.00 mol C12.01 g C

1.00 mol Ti2.00 mol C

1.19 x 103

1000 g TiO21 kg TiO2

1.00 mol TiO279.87 g TiO2

1.00 mol Ti1.00 mol TiO2

theoretical yield

percent yield


percent yield.

TiO2(s) + 2 C(s)➜Ti(s) + 2 CO(g)

1.10 x 103 mol Ti x x = 52.9 kg Ti 47.87 g Ti1 mol Ti

1.00 kg Ti1000 g Ti

7. How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of copper(II) oxide?

2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)

If 4.61 g of N2 are made, what is the percent yield?

smaller amount is from limiting reactant}

molNH3

g

mol

molN2

mol

mol

gNH3

gCuO

molCuO

molN2

45.2 g CuO x x = 0.1894 mol N2

Limiting reactant Smaller # moles of N2

Theoretical yield

9.05 g NH3 x x = 0.2657 mol N2 1.00 mol NH317.03 g NH3

1.00 mol N22.00 mol NH3

1.00 mol CuO79.55 g CuO

1.00 mol N23.00 mol CuO

How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of copper(II) oxide?

2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 are made, what is the percent yield?


2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 are made, what is the percent yield?

Theoretical Yield

smaller molN2

gN2

Theoretical Yield

Actual Yield

= % Yield

0.189 mol N2 x = 5.30 g N228.02 g N2

1.00 mol N2

0.1894 mol N2 x = 5.307 g N228.02 g N2

1.00 mol N2


2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)

If 4.61 g of N2 are isolated, what is the percent yield?

45.2 g CuO x x = 0.1894 mol N21.00 mol CuO79.55 g CuO

1.00 mol N23.00 mol CuO

Theoretical yield

percent yield

4.61 g N25.30 g N2

x 100% = 87.0 % 4.61 g N25.31 g N2

x 100% = 86.8 %

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Chapter 3 Calculations Related to Chemical Formulas and ...profkatz.com/courses/wp-content/uploads/2017/01/CH1710-Lecture-7... · Chapter 3 Calculations Related to Chemical Formulas