Question

Express the composition of each of the following compounds as the mass percents of its elements. a. formaldehyde, \(\mathrm{CH}_{2} \mathrm{O}\) b. glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) c. acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

Short answer

The mass percents of the elements in the compounds are as follows: a. Formaldehyde: 39.98% carbon, 6.71% hydrogen, and 53.31% oxygen. b. Glucose: 40.00% carbon, 6.72% hydrogen, and 53.28% oxygen. c. Acetic acid: 6.73% hydrogen, 40.02% carbon, and 53.25% oxygen.

Step-by-step solution

a. Formaldehyde, \(\mathrm{CH}_{2} \mathrm{O}\)#

To find the mass percent of each element in formaldehyde, we must first find the molar mass of the compound. To determine the molar mass of formaldehyde, we need to know the molar masses of its constituent elements, which are carbon (C), hydrogen (H), and oxygen (O). The molar masses are roughly: - Carbon (C): 12.01 g/mol - Hydrogen (H): 1.01 g/mol - Oxygen (O): 16.00 g/mol Now, we can calculate the molar mass of formaldehyde by summing up the molar masses of its elements: \[ (1 \times 12.01) + (2 \times 1.01) + (1 \times 16.00) = 30.03 \ g/mol \] Next, we will find the mass percent of each element in formaldehyde: - Mass percent of carbon: \[ \frac{1 \times 12.01}{30.03} \times 100 = 39.98\% \] - Mass percent of hydrogen: \[ \frac{(2 \times 1.01)}{30.03} \times 100 = 6.71\% \] - Mass percent of oxygen: \[ \frac{1 \times 16.00}{30.03} \times 100 = 53.31\% \] Thus, the mass percents in formaldehyde are approximately 39.98% carbon, 6.71% hydrogen, and 53.31% oxygen.

b. Glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)#

To find the mass percent of each element in glucose, we must first find the molar mass of the compound: \[ (6 \times 12.01) + (12 \times 1.01) + (6 \times 16.00) = 180.18 \ g/mol \] Next, we will find the mass percent of each element in glucose: - Mass percent of carbon: \[ \frac{(6 \times 12.01)}{180.18} \times 100 = 40.00\% \] - Mass percent of hydrogen: \[ \frac{(12 \times 1.01)}{180.18} \times 100 = 6.72\% \] - Mass percent of oxygen: \[ \frac{(6 \times 16.00)}{180.18} \times 100 = 53.28\% \] Thus, the mass percents in glucose are approximately 40.00% carbon, 6.72% hydrogen, and 53.28% oxygen.

c. Acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)#

To find the mass percent of each element in acetic acid, we must first find the molar mass of the compound: \[ (1 \times 1.01) + (2 \times 12.01) + (3 \times 1.01) + (2 \times 16.00) = 60.05 \ g/mol \] Next, we will find the mass percent of each element in acetic acid: - Mass percent of hydrogen: \[ \frac{(4 \times 1.01)}{60.05} \times 100 = 6.73\% \] - Mass percent of carbon: \[ \frac{(2 \times 12.01)}{60.05} \times 100 = 40.02\% \] - Mass percent of oxygen: \[ \frac{(2 \times 16.00)}{60.05} \times 100 = 53.25\% \] Thus, the mass percents in acetic acid are approximately 6.73% hydrogen, 40.02% carbon, and 53.25% oxygen.