Terms to Know

Percent composition – relative amounts of each element in a compound

Empirical formula – lowest whole- number ratio of the atoms of an element in a compound

An 8.20 g piece of magnesium combines completely with 5.40 g of oxygen to form a compound. What is the percent composition of this compound?

1. Calculate the total mass

2. Divide each given by the total mass

and then multiply by 100%

3. Check your answer: The

percentages should total 100%

Answer

The total mass is 8.20 g + 5.40 g = 13.60 g

Divide 8.2 g by 13.6 g and then multiply by 100% = 60.29412 = 60.3%

Divide 5.4 g by 13.6 g and then multiply by 100% = 39.70588 = 39.7%

Check your answer: 60.3% + 39.7% = 100%

Calculate the percent composition of propane (C3H8)

1. List the elements2. Count the atoms3. Multiply the number of atoms of the element by the atomic mass of the element (atomic mass is on the periodic table)4. Express each element as a percentage of the total molar mass5. Check your answer

Answer

Total molar mass = 44.0 g/mol

36.0 g C = 81.8%

8.0 g H = 18.2%

Calculate the mass of carbon in 52.0 g of propane (C3H8)

1. Calculate the percent composition using the formula (See previous problem)

2. Determine 81.8% of 82.0 g

Move decimal two places to the

left (.818 x 82 g)

3. Answer = 67.1 g

Calculating Empirical Formulas

Microscopic – atoms

Macroscopic – moles of atoms

Lowest whole-number ratio may not be the same as the compound formula

Example: The empirical formula of hydrogen peroxide (H2O2) is HO

Empirical Formulas

The first step is to find the mole-to-mole ratio of the elements in the compoundIf the numbers are both whole numbers, these will be the subscripts of the elements in the formulaIf the whole numbers are identical, substitute the number 1

Example: C2H2 and C8H8 have an empirical formula of CHIf either or both numbers are not whole numbers, numbers in the ratio must be multiplied by the same number to yield whole number subscripts

What is the empirical formula of a compound that is 25.9% nitrogen

and 74.1% oxygen?

1. Assume 100 g of the compound, so that

there are 25.9 g N and 74.1 g O

2. Convert to mole-to-mole ratio:

Divide each by mass of one mole

25.9 g divided by 14.0 g = 1.85 mol N

74.1 g divided by 16.0 g = 4.63 mol O

3. Divide both molar quantities by the

smaller number of moles

4. 1.85/1.85 = 1 mol N 4.63/1.85 = 2.5 mol O

5. Multiply by a number that converts each to a whole number (In this case, the number is 2 because 2 x 2.5 = 5, which is the smallest whole number )2 x 1 mol N = 22 x 2.5 mol O = 5Answer: The empirical formula is N2O5

Determine the Empirical Formulas

1. H2O2

2. CO2

3. N2H4

4. C6H12O6

5. What is the empirical formula of a compound that is 3.7% H, 44.4% C, and 51.9% N?

Answers

Compound Empirical Formula

1. H2O2 HO

2. CO2 CO2

3. N2H4 NH2

4. C6H12O6 CH2O

5. HCN

Calculating Molecular FormulasThe molar mass of a compound is a simple whole-number multiple of the molar mass of the empirical formula

The molecular formula may or may not be the same as the empirical formula

Calculate the molecular formula of the compound whose molar mass is 60.0 g and empirical formula is CH4N.

1. Using the empirical formula, calculate the empirical formula mass (efm)

(Use the same procedure used to calculate molar mass.)2. Divide the known molar mass by the efm 3. Multiply the formula subscripts by this value to get the molecular formula

Answer

Molar mass (efm) is 30.0 g

60.0 g divided by 30.0 g = 2

Answer: C2H8N2

Practice Problems1) What is the empirical formula of a compounds that is 25.9% nitrogen and 74.1% oxygen?

2) Calculate the empirical formula of a compound that is 32.00% C, 42.66% O, 18.67% N, and 6.67% H.

3) Calculate the empirical formula of a compound that is 42.9% C and 57.1% O.

Practice Problems4) What is the molecular formula for each compound:

a) CH2O: 90 g

b) HgCl: 472.2 g

c) C3H5O2: 146 g