Math in Chemistry

Percent Composition

Purpose: Can be used to figure out chemical formulas.

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

Masses are Given Examples

1) A sample of Silver sulfide was found to contain 29.0 g of Ag and 4.30 g of S. Calculate the percent composition.

Masses are Given Examples

2) 222.6 g of Na combines completely with 77.4 g of O. Calculate the percent composition.

No Masses Given

Steps to solve problem: 1) Assume you have 1 mole of the compound and

calculate its molar mass 2) Determine the molar masses of each element

in the compound 3) Divide the molar mass for the element by the

molar mass of the compound 4) Multiply by 100 to get the percent

No Masses GivenExamples

1) Calculate the percent composition of C and H in ethane, C2H6.

No Masses GivenExamples

2) Calculate the percent composition of sodium hydrogen sulfate.

II. Calculate the Mass of an Element in a Compound

Steps to solve the problem: 1) Find the molar mass of the compound 2) Find the percent composition of the element 3) Set up a conversion factor problem

Given mass (g compound) x percent composition/100 (g compound)

Calculate the Mass of an Element in a Compound

Examples 1) Calculate the mass of hydrogen in the

following: A) 350 g C3H8

B) 20.2 g NaHCO3

C) 378 g HCN

III. Empirical Formulas

Empirical formulas are the lowest whole number ratio of the atoms of the elements in a compound.

Molecular Formula C2O4 Na2O2

Empirical Formula CO2 NaO

III. Empirical Formulas

Steps to solve a problem: 1) If the problem is given in percents, assume 100

g of the compound. This lets you easily convert the percents to grams (10% of 100g = 10g)

2) Convert the mass of each element to moles of that element

3) Find the smallest whole number ratio between the moles of the elements by dividing each molar mass by the smallest molar mass present.

4) The numbers you get as answers tell how many atoms of that element are present in the compound. If the numbers do not come out whole, round to the nearest whole number.

Empirical FormulasExamples

1) What is the empirical formula of a compound that is 49.6% nitrogen and 50.4% oxygen?

Empirical FormulasExamples

2) What is the empirical formula for a compound that is 79.8% C and 20.2% H?

Empirical FormulasExamples

3) What is the empirical formula of a compound that is 67.6% Hg, 10.8% S, and 21.6% O?

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

IV. Molecular Formulas

Steps to solve the problem: 1) Determine the empirical formula 2) Divide the molecular mass by the empirical

formula mass to get a ratio. 3) Multiply the elements’ subscripts by the

number you get in step 2.

Molecular FormulasExamples

1) Calculate the molecular formulas of the following compounds:

Molecular Mass Empirical Formula 60 g CH4N 78 g NaO 181.5 g C2HCl

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 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 34.0 g/mol, what is the molecular formula?