Stoichiometry Worksheet (Page 3 of 6) in pdf

Example: Calculate the percent composition of ammonium nitrate.

Description of Action

Action

1. Write the formula for the given compound.

1. NH

2. Record the amount of each element in the compound.

2. N: 2 x 14.0 = 28.0

(Note: We have 2 total nitrogen, so we record them

H: 4 x 1.0 = 4.0

together.) Multiply the amount of each element by its

O: 3 x 16.0 = 48.0

atomic weight (measured to the tenths place). Add the

80.0 g/mol

results to find the gram formula mass of the compound.

3. Divide the total mass of each element by the gram

3. N: 28.0 ÷ 80.0 = 0.350

formula mass. For these calculations your answer should

H: 4.0 ÷ 80.0 = 0.050

have 3 places after the decimal (round if necessary).

O: 48.0 ÷ 80.0 = 0.600

4. Multiply each result by 100. Add the % symbol to your

4. N: 0.350 x 100 = 35.0%

new result. (If you were to add up your percentages the

H: 0.050 x 100 = 5.0%

must equal 100.)

O: 0.600 x 100 = 60.0%

100.0%

Determining Empirical and Molecular Formula (LO 1.2)

Empirical & Molecular Formula

Empirical Formula

Given: Percentage or mass of each element or compound.

1. Divide each percentage or mass given by the molar mass of the element or compound.

2. Divide each result by the smallest result.

3. Round and multiply (if necessary) each result by the SAME whole number to get a whole

number result. (This step is not necessary for hydrates). Use the following chart:

Round to:

Multiply by:

x.00

(use the other factor)

x.20

x.25

x.33

x.50

x.66

x.75

x.80

4. Write the formula using your results.

Molecular Formula (do all of the above)

5. Find the gram formula mass of the empirical formula.

6. Divide the molecular formula mass (given) by the empirical formula mass (calculated).

7. Multiply each subscript by the result. (The result MUST be a whole number.)

Example: Ascorbic acid, also known as vitamin C, has a percentage composition of 40.9% carbon, 4.58% hydrogen,

and 54.5% oxygen. Its molecular mass is 176.1 g/mol. What is its molecular formula?

C: 40.9 ÷ 12.0 = 3.41 ÷ 3.41 = 1.00 x 3 = 3

C: 3 x 12.0 = 36.0

176.1 ÷ 88.0 = 2

H: 4.58 ÷ 1.0 = 4.58 ÷ 3.41 = 1.32 x 3 = 4

H: 4 x 1.0 = 4.0

O: 54.5 ÷ 16.0 = 3.41 ÷ 3.41 = 1.00 x 3 = 3

O: 3 x 16.0 = 48.0

(empirical formula)

88.0 g/mol

Stoichiometry Worksheet Page 3

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