Tis the season to be thankful so lets thank Avogadro for math in

Chemistry

Percent Composition

Purpose: Can be used to figure out chemical formulas. the percentage by mass of each element in a

compound

100mass total

element of massncompositio %

I. Percent Composition

Two different types of problems: 1) Masses are given 2) No Masses are given

Masses are Given

Steps to solve problem: 1) Add given masses to get total mass for one

compound 2) Divide mass of each element by the total mass 3) Multiply by 100 to get the percent

%Fe =28 g

36 g 100 =78% Fe

%O =8.0 g

36 g 100 =22% O

Find the percentage composition of a sample that is 28 g Fe and 8.0 g O.

A. Percentage Composition

No Masses Given

Steps to solve problem: 1) Assume you have 1 mole of the compound 2) Calculate the molar mass of each element in

the compound by multiplying the subscript by the molar mass of the element

3) Divide the molar mass for the element by the total molar mass of the compound

4) Multiply by 100 to get the percent

No Masses GivenExamples

1) Calculate the percent composition of oxygen in water.

H2O H: 2 x 1.008 = 2.016 g/mol O: 1 x 16.00 = 16.00 g/mol Total molar mass (2.016 + 16.00) = 18.02

g/mol %O = (16.00 / 18.02) x 100 = 88.79%

No Masses GivenExamples

2) Calculate the percent composition of calcium carbonate.

100.09 g/mol %Ca = 40.08g/100.09g×100 = 40.04% %C = 12.01g/100.09g×100 =12.00% %O = 3(16.00g)/100.09g×100 = 47.96%

B. Empirical Formula

C2H6

CH3

reduce subscripts

Smallest whole number ratio of atoms in a compound

B. Empirical Formula1. Find mass (or %) of each element.

2. Find moles of each element. (divide given mass by molar mass)

3. Divide answers by the smallest # to find subscripts.

4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

Remember!

Percent to MassMass to MoleDivide by SmallMultiply til Whole

B. Empirical FormulaFind the empirical formula for a

sample of 25.9% N and 74.1% O.

25.9 g 1 mol

14.01 g = 1.85 mol N

74.1 g 1 mol

16.00g = 4.63 mol O

1.85 mol

1.85 mol

= 1 N

= 2.5 O

B. Empirical Formula

N1O2.5

Need to make the subscripts whole numbers multiply by 2

N2O5If .0_ something or .9_ something, just round

Empirical FormulasExamples

3) Ascorbic acid (vitamin C) contains C (40.89%), H (4.56%), and O (54.55%) by mass. What is the empirical formula of ascorbic acid?

C3H4O3

C. Molecular Formula“True Formula” - the actual number

of atoms in a compound

CH3

C2H6

empiricalformula

molecularformula

?

IV. Molecular Formulas

Usually the empirical formula is the molecular formula for a compound. When it is not, the molecular formula is defined as the elements and number of atoms that are contained in a compound.

IV. Molecular Formulas

Molecular formulas are always multiples of empirical formulas.

CH3 C2H6

C. Molecular Formula1. Find the empirical formula.2. Find the empirical formula mass.3. Divide the molecular mass by the empirical mass.4. Multiply each subscript by the answer from step 3.

nmass EF

mass MF nEF

C. Molecular FormulaThe empirical formula for ethylene is CH2. Find the molecular

formula if the molecular mass is 28.0532 g/mol?

28.0532 g/mol

14.03 g/mol = 2.000

empirical mass = 14.03 g/mol

(CH2)2 C2H4

Molecular FormulasExamples

2) The compound methyl butanoate smells like apples. Its percent composition is 58.8% C, 9.8% H, and 31.4% O. If the molecular mass is 102.1317 g/mol, what is the molecular formula?

Molecular FormulasExamples

3) You find 7.36 g of a compound has decomposed to give 6.93 g of oxygen. The rest is hydrogen. If the molecular mass is 34.0147 g/mol, what is the molecular formula?