Chapter 3
Q. 3.6
The compound that gives vinegar its sour taste is acetic acid, which contains the elements carbon, hydrogen, and oxygen.
When 5.00 g of acetic acid are burned in air, 7.33 g of CO_{2} and 3.00 g of water are obtained. What is the simplest formula of acetic acid?
| ANALYSIS | |
| elements in acetic acid (C, H, and O) mass of acetic acid (5.00 g) result of combustion analysis (7.33 g CO_{2}, 3.00 g H_{2}O) |
Information given: |
| molar masses (MM) of CO_{2} and H_{2}O | Information implied: |
| simplest formula of acetic acid | Asked for: |
STRATEGY
1. Find the mass of C and H by using the conversion factors:
\frac{12.01 g C}{44.01 g CO_{2}} \frac{2(1.008) g H}{18.02 g H_{2}O}
2. Find the mass of O by difference:
mass of sample = mass of C + mass of H + mass of O
3. Follow the schematic pathway shown in Figure 3.5.
Step-by-Step
Verified Solution
| 7.33 g CO_{2} × \frac{12.01 g C}{44.01 g CO_2} = 2.00 g C
3.00 g H_{2}O × \frac{2(1.008) g H}{18.02 g H_{2}O} = 0.336 g H |
1. mass of C
mass of H |
| mass of O = mass of sample – (mass of C + mass of H) = 5.00 g – (2.00 g + 0.336 g) = 2.66 g |
2. mass of O |
| C: \frac{2.00 g C}{12.01 g/mol} = 0.167; H: \frac{0.336 g H}{1.008 g/mol} = 0.333; O: \frac{2.66 g O}{16.00 g/mol} = 0.166
C: \frac{0.167 mol}{0.166 mol} = 1; H: \frac{0.333 mol}{0.167 mol} = 2; O: \frac{0.166 mol}{0.166 mol} = 1 Using the ratios as subscripts, the simplest formula is CH_{2}O. |
3. mol of each element
ratios simplest formula |