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## Q. 5.4

GRADED

Acetone is widely used as a nail polish remover. A sample of liquid acetone is placed in a 3.00-L flask and vaporized by heating to 95°C at 1.02 atm. The vapor filling the flask at this temperature and pressure weighs 5.87 g.

ⓐ What is the density of acetone vapor under these conditions?
ⓑ Calculate the molar mass of acetone.
ⓒ Acetone contains the three elements, C, H, and O. When 1.000 g of acetone is burned, 2.27 g of $CO_{2}$ and 0.932 g of $H_{2}O$ are formed. What is the molecular formula of acetone?

 ANALYSIS volume of the flask (3.00 L); mass of acetone vapor (5.87 g) Information given: volume of the vapor Information implied: density of acetone vapor Asked for:

STRATEGY

1. Recall the formula for density (density = mass/volume).
2. A gas occupies the volume of the flask. Volume of vapor = volume of flask

 ANALYSIS volume of the flask (3.00 L); mass of acetone vapor (5.87 g); pressure (P) (1.02 atm); temperature (T) (95°C) Information given: molar mass of acetone Asked for:

STRATEGY

1. To find molar mass you need to know mass and n (molar mass = mass/n). Mass is given.
2. Use the ideal gas law to find n (n = PV/RT).

 ANALYSIS from part (b), molar mass of acetone (58.1 g/mol) The combustion of 1.00 g of sample yields 2.27 g $CO_{2}$ and 0.932 g $H_{2}O$. Information given: mass of C, H, and O in 1.00-g sample Information implied: molecular formula of acetone Asked for:

STRATEGY

1. Recall from Section 3.2 how to convert the mass of the product of combustion to the mass of the element.
2. Follow Figure 3.5 to obtain the simplest formula for the compound.
3. Compare the simplest formula’s molar mass to the molar mass obtained in part (b).

## Verified Solution

 density = $\frac{mass}{volume}$ = $\frac{5.87 g}{3.00 L}$ = 1.96 g/L density

 n = $\frac{PV}{RT}$ = $\frac{(1.02 atm × 3.00 L)}{(95 + 273)K × 0.0821(L · atm/mol · K)}$ = 0.101 mol moles (n) molar mass = $\frac{mass}{n}$ = $\frac{5.87 g}{0.101 mol}$ = 58.1 g/mol molar mass

 mass C: 2.27 g $CO_{2}$ × $\frac{12.01 g C}{44.01 g CO_{2}}$=0.619 g mass H: 0.932 g $H_{2}O$ × $\frac{2(1.008) g H}{18.02 g H_{2}O}$ = 0.104 g mass O = mass sample – (mass C 1 mass H) = 1.000 g – (0.619 + 0.104) g = 0.277 g Mass of each element C: $\frac{0.619 g}{12.01 g/mol}$ = 0.0515 mol; H: $\frac{0.104 g}{1.008 g/mol}$ = 0.103 mol; O:$\frac{0.277 g}{16.00 g/mol}$ = 0.0173 mol moles of each element C: $\frac{0.0515}{0.0173}$ = 3; H: $\frac{0.104}{0.0173}$ = 6; O: $\frac{0.0173}{0.0173}$ = 1 Atomic ratios $C_{3}H_{6}O$ Simplest formula 3(12.01) + 6(1.008) + 16.00 = 58.08 g/mol MM of simplest formula 58.1 g/mol MM of vapor (from part (b)) $C_{3}H_{6}O$ (simplest formula 5 molecular formula) Molecular formula