Question 8.10: Butane, C4H10, is used as a fuel for camping stoves. If you ...
Butane, C_{4}H_{10}, is used as a fuel for camping stoves. If you have 108 mL of butane gas at 715 mmHg and 25 °C, what is the mass, in grams, of butane?
STEP 1 State the given and needed quantities .When three of the quantities (P, V, and T) are known, we use the ideal gas law equation to solve for the unknown quantity, moles (n). Because the pressure is given in mmHg,
we will use R in mmHg. The volume given in milliliters (mL) is converted to a volume in liters (L). The temperature is converted from degrees Celsius to kelvins.
| ANALYZE THE PROBLEM | Given | Need | Connect |
| P = 715 mmHg
V = 108 mL(0.108 L) T = 25 °C + 273 = 298 K |
\boxed{n} | ideal gas law, PV = nRT, molar mass |
STEP 2 Rearrange the ideal gas law equation to solve for the needed quantity. By dividing both sides of the ideal gas law equation by RT, we solve for moles, n.
PV=\boxed{n}RT Ideal gas law equation
\frac{PV}{RT}=\frac{\boxed{n } \cancel{RT}}{\cancel{RT}}
\boxed{n} =\frac{PV}{RT}
STEP 3 Substitute the gas data into the equation, and calculate the needed quantity.
\boxed{n} =\frac{715 \cancel{mmHg} \times 0.108 \cancel{L}}{\frac{62.4 \cancel{L} \cdot \cancel{mmHg}}{mole \cdot \cancel{K}} \times 298 \cancel{K}} = 0.00415 mole(4.15 \times 10^{-3} mole)Now we can convert the moles of butane to grams using its molar mass of 58.12 g/mole:
0.00415 \cancel{mole C_{4}H_{10}} \times \frac{58.12 g C_{4}H_{10}}{1 \cancel{mole C_{4}H_{10}}} = 0.241 g of C_{4}H_{10}