chemistry (stoichiometry)

Question

Cinnamic acid (monoprotic) contains only carbon, hydrogen and oxygen, and is found by analysis to be

73.0% C and 5.4% hydrogen. In a titration, 18.02 mL of 0.135M NaOH is found to neutralize 0.3602 g of

cinnamic acid.

a. Find the empirical formula of this compound.

b. Find the molar mass of this compound.

c. Write the molecular formula for this compound.

Answer 1

A. To calculate the empirical formula, we must determine the relative molar ratios of the elements within the compound. From the experimental data, we know the mass percent of each element. This means, in a 100 gram sample, we have:

73.0 grams of carbon

5.4 grams of hydrogen

21.6 grams of oxygen (100% - 73.0% - 5.4%)

From here, we can convert the grams into moles by dividing by each element's molar mass.

73.0 g C / 12.011 g/mol = 6.08 mol C

5.4 g H / 1.008 g/mol = 5.36 mol H

21.6 g O / 15.999 g/mol = 1.35 mol O

To figure out the molar ratios, we divide by the least number of moles. In this case, oxygen only has 1.35 moles.

6.08 mol / 1.35 mol = 4.5 C

5.36 mol / 1.35 mol = 3.97 H (round to 4.0)

1.35 mol / 1.35 mol = 1.00 O

We only want integer numbers. The 3.97 we can safely approximate to be about 4.0. The 4.5 can be turned into an integer by multiplying all numbers by 2.

4.5 C -> 9.0 C

4.0 H -> 8.0 H

1.0 O -> 2.0 O

Now we have our empirical formula: C₉H₈O₂

B. To calculate the molar mass, we need to figure out how many moles are in the 0.3602 g that were used in the neutralization reaction. Because cinnamic acid is monoprotic, it will react with NaOH in a 1:1 molar ratio. Therefore, the number of moles of NaOH and cinnamic acid used in the titration will be the same.

The number of moles of NaOH can be calculated by multiplying the concentration of NaOH with the volume used.

0.135 M * 0.01802 L = 0.00243 moles NaOH (note the unit change from mL to L)

If 0.00243 moles of NaOH was used in the titration, then we must have had 0.00243 moles of cinnamic acid as well.

Now we can calculate the molar mass of cinnamic acid using the formula:

Molar mass = mass / moles

MM = 0.3602 g / 0.00243 moles

MM = 148 g/mol

C. We know the empirical formula and the molar mass. The molar mass must be a multiple of the empirical mass. So first, let's calculate the empirical mass.

C₉H₈O₂ = 9x12.011 + 8x1.008 + 2x15.999 = 148.2 g/mol

Since the empirical mass and the molar mass are the same (within rounding), the empirical formula and the molecular formula must also be the same. Therefore, the molecular formula is: C₉H₈O₂