Section 3 Chemical Formulas and Equations. 2 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.;

Section 3

Chemical Formulas and Equations

2Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General

Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

Mass and Moles of a Substance

• Chemistry requires a method for determining the numbers of molecules in a given mass of a substance.– This allows the chemist to carry out “recipes” for

compounds based on the relative numbers of atoms involved.

– stoichiometry - the calculation involving the quantities of reactants and products in a chemical equation.



Molecular Weight and Formula Weight

• The molecular weight (covalent bonds/nm-nm) of a substance is the sum of the atomic weights of all the atoms in a molecule of the substance.

– For, example, a molecule of H2O contains 2 hydrogen atoms (at 1.01 amu each) and 1 oxygen atom (16.00amu), giving a molecular weight of 18.02 amu.



Molecular Weight and Formula Weight

• The formula weight of a substance is the sum of the atomic weights of all the atoms in one formula unit of the compound, whether molecular or not.– For example, one formula unit (FU) of NaCl

contains 1 sodium atom (22.99 amu) and one chlorine atom (35.45 amu), giving a formula weight of 58.44 amu.



Mass and Moles of a Substance• The Mole Concept

– A mole is defined as the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 grams of carbon–12.

– The number of atoms in a 12-gram sample of carbon–12 is called Avogadro’s number (Na). The value of Avogadro’s number is 6.022 x 1023.

1 mole = 6.022 x 1023 ? ions, particles, atoms, molecules, items, etc.

1 mole Na2CO3 6.022 x 1023 FU Na2CO3

1 mole CO2 6.022 x 1023 molecules CO2



Mass and Moles of a Substance• The molar mass (Mm) of a substance is the

mass of one mole of a substance.– For all substances, molar mass, in grams per

mole, is numerically equal to the formula weight in atomic mass units.

– That is, one mole of any element weighs its atomic mass in grams.

1 molecule of H2O - MW = 18.02 amu

1 mole of H2O - Mm = 18.02 g H2O

1 formula unit of NaCl - FW = 58.44 amu

1 mole of NaCl - Mm = 58.44 g NaCl



• How is it possible that Mm and FW/MW are the same value but different units?

• A.)What is the mass in grams of one Cl atom? B.)in one HCl molecule?




• Mole calculations– Converting the number of moles of a given substance

into its mass, and vice versa, is fundamental to understanding the quantitative nature of chemical equations.

58.44 g NaCl 1 mole NaCl

1 mole NaCl 58.44 g NaCl

A"" of mass molecular)(or atomicA"" of mass

A"" of moles




• Mole calculations– Suppose we have 5.75 moles of magnesium. What

is its mass?




• Mole calculations– Suppose we have 100.0 grams of H2O. How many

moles does this represent?



Mass and Moles and Number of Molecules or Atoms

• The number of molecules or atoms in a sample is related to the moles of the substance:

atomsFe1002.6Femole1

moleculesHCl10 6.02 HCl mole 123

23

• How many molecules are there in 56 mg HCN?

HW 20-22



Determining Chemical Formulas• The percent composition of a compound is the mass

percentage of each element in the compound.

– We define the mass percentage of “A” as the parts of “A” per hundred parts of the total, by mass. That is,

%100whole the of mass

whole in A"" of mass A"" % mass

sample g 100

A g 20 A % 20



Mass Percentages from Formulas

• Let’s calculate the percent composition (%C, %H) of one molecule of butane, C4H10.

First, we need the molecular wt of C4H10.

amu 48.04 amu/atom 12.01 @ carbons 4 amu 10.10 amu/atom 1.01 @ hydrogens 10 amu 58.14 HC of molecule 1 104

Now, we can calculate the percents.

C%63.82%100 C % amu total 14.58Camu 48.04

H%37.17%100 H % amu total 14.58Hamu 10.10



• How many grams of carbon are there in 83.5 g of formaldehyde, CH2O, (40.0% C, 6.73% H, 53.3% O)?



• An unknown acid contains only C, H, O. A 4.24 mg sample of acid is completely burned. It gives 6.21 mg of CO2 and 2.54 mg H2O. What is mass% of each element in the unknown acid?



Determining Chemical Formulas

• Determining the formula of a compound from the percent composition.– The percent composition of a compound leads

directly to its empirical formula.– An empirical formula (or simplest formula) for a

compound is the formula of the substance written with the smallest integer (whole number) subscripts.




• Determining the empirical formula from the percent composition.– Benzoic acid is a white, crystalline powder used as

a food preservative. The compound contains 68.8% C, 5.0% H, and 26.2% O by mass. What is its empirical formula?

– In other words, give the smallest whole-number ratio of the subscripts in the formula

Cx HyOz



Determining Chemical Formulas– For the purposes of this calculation, we will

assume we have 100.0 grams of benzoic acid.

– Then the mass of each element equals the numerical value of the percentage.

– Since x, y, and z in our formula represent mole-mole ratios, we must first convert these masses to moles.

Cx HyOz




C )8(72.5g 12.01

C mol 1 8.68 molCg

H )5(9.4g 1.01

H mol 1 0.5 molHg

O )7(63.1g 16.00

O mol 1 2.26 molOg

This isn’t quite a whole number ratio, but if we divide each number by the smallest of the three, a better ratio might emerge.

• Determining the empirical formula from the percent composition.– Our 100.0 grams of benzoic acid would contain:




• Determining the empirical formula from the percent composition.– Our 100.0 grams of benzoic acid would contain:

3.501.63(7)C 728.5 mol

3.01.63(7)H .954 mol

1.001.63(7)O )7(63.1 mol

now it’s not too difficult to see that the smallest whole number ratio is 7:6:2 (multiple everything by 2 to get whole number.The empirical formula is C7H6O2 .2 x C3.5H3O1 C7H6O2



Determining Chemical Formulas– An empirical formula gives only the smallest whole-number ratio of

atoms in a formula.

– The molecular formula should be a multiple of the empirical formula (since both have the same percent composition).

C2H3O2 empirical formula (lowest whole #)

C4H6O4 molecular formula

C8H12O8 molecular formula

Which is not an empirical formula?

CH4 CH4O C2H4O2 C2H6O

– To determine the molecular formula, we must know the molecular weight (molar mass) of the compound.



Determining Chemical Formulas• Determining the molecular formula from the

empirical formula.

Molecular weight = n x empirical formula wt. where n is the multiple

factor

n = molecular wt

empirical wt



• Acetic Acid contains 39.9% C, 6.7% H, 53.4% O. Determine empirical formula. The molecular mass of acetic acid is 60.0 g/mol. What is the molecular formula?

HW 23



Stoichiometry: Quantitative Relations in Chemical Reactions

• Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction.– It is based on the balanced chemical equation and

on the relationship between mass and moles.

– Such calculations are fundamental to most quantitative work in chemistry.



28.02 g N2 + 3(2.02 g) H2 2 (17.04 g) NH3

28.02 g + 6.06 g 34.08 g

34.08 g = 34.08 g

)g(NH 2 (g)3H (g)N 322

322 NH mol 2 H mol 3 N mol 1 1 molecule N2 + 3 molecules H2 2 molecules NH3

Molar Interpretation of a Chemical Equation

• The balanced chemical equation can be interpreted in numbers of molecules, but generally chemists interpret equations as “mole-to-mole” relationships.



Molar Interpretation of a Chemical Equation

• Suppose we wished to determine the number of moles of NH3 we could obtain from 4.8 mol H2 (assume N2 in excess).

)g(NH 2 (g)3H (g)N 322



Mass Relationships in Chemical Equations

How many grams of HCl are required to react with 5.00 grams MnO2 according to this equation?

)s(MnO HCl(aq) 4 2 )g(Cl (aq)MnCl O(l)H 2 222



Mass Relationships in Chemical Equations

How many grams of CO2 gas can be produced from 1.00 kg Fe2O3?

Fe2O3 (s) + 3 CO (g) 2 Fe (s) + 3 CO2 (g)

HW 24



Limiting Reagent• The limiting reactant (or limiting reagent) is the reactant

that is entirely consumed when the reaction goes to completion.

• The limiting reagent ultimately determines how much product can be obtained.– For example, bicycles require one frame and two wheels. If you have

20 wheels but only 5 frames, how many bicycles can be made?

o o o o o o o o o o o o o o o o o o o o 20 wheels

1 frame + 2 wheels 1 bike



Limiting Reagent

If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many moles of H2 are produced?

)g(H (aq)ZnCl HCl(aq) 2 Zn(s) 22



If 7.36 g Zn was heated with 6.45 g sulfur, what amount of ZnS was produced?

ZnS8 S Zn(s)8 8



Theoretical and Percent Yield

• The theoretical yield of product is the maximum amount of product that can be obtained from given amounts of reactants.

– The percentage yield is the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated).

%100(calc) yield ltheoretica

(exp) yield actual Yield%



Theoretical and Percent Yield

• To illustrate the calculation of percentage yield, recall that the theoretical yield of ZnS in the previous example was 11.0 g ZnS.

• If the actual yield of the reaction had been 9.32 g ZnS, what is the %yield?

%7.84%100 ZnSg 11.0

ZnSg 9.32 Yield%

HW 25



If 11.0 g CH3OH are mixed with 10.0 g CO, what is the theoretical yield of HC2H3O2 in the following reaction? If the actual yield was

19.1 g, what is the %yield of HC2H3O2?

CH3OH (l) + CO (g) HC2H3O2 (l)

Documents

Section 3 Chemical Formulas and Equations. 2 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.;