Question

Propane gas, \(\mathrm{C}_{3} \mathrm{H}_{8}\), a hydrocarbon, is used as a fuel for many barbecues. a. How many grams of the compound are in \(1.50\) moles of propane? b. How many moles of the compound are in \(34.0 \mathrm{~g}\) of propane? c. How many grams of carbon are in \(34.0 \mathrm{~g}\) of propane?

Short answer

a. 66.165 g b. 0.771 moles c. 27.77 g

Step-by-step solution

- Calculate the molar mass of propane (C\textsubscript{3}H\textsubscript{8})

First, sum the molar masses of all the atoms in the propane molecule. Propane has 3 carbon atoms and 8 hydrogen atoms. The molar mass of carbon is approximately 12.01 g/mol and that of hydrogen is approximately 1.01 g/mol. \[\text{Molar mass of } \text{C}_3\text{H}_8 = (3 \times 12.01 \text{ g/mol}) + (8 \times 1.01 \text{ g/mol}) = 36.03 \text{ g/mol} + 8.08 \text{ g/mol} = 44.11 \text{ g/mol} \]

- Calculate the mass of propane in 1.50 moles (Part a)

Using the molar mass calculated in Step 1, multiply it by the number of moles. \[\text{Mass} = \text{moles} \times \text{molar mass} = 1.50 \text{ moles } \times 44.11 \text{ g/mol } = 66.165 \text{ g} \]

- Calculate the number of moles in 34.0 grams of propane (Part b)

Using the molar mass from Step 1, divide the given mass by the molar mass to find out the number of moles. \[\text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{34.0 \text{ g}}{44.11 \text{ g/mol}} \ \text{Moles} \begin{aligned} &\text{=} \boxed{\text{0.771 moles}} \ \text{0.771 moles} \ \text{} \ \text = \text{0.771} \ \text{} \ \text \frac{}{} \text{} \]

- Calculate the mass of carbon in 34.0 grams of propane (Part c)

First, determine the fraction of the total mass of propane that is due to carbon. This is done by calculating what portion of the molar mass is from carbon. For three carbon atoms, the mass contribution is: \[(3 \times 12.01 \text{ g/mol}) = 36.03 \text{ g/mol} \] Then, calculate the mass of carbon in 34.0 grams of propane. \[\text{Mass of carbon} = \frac{36.03 \text{ g/mol}}{44.11 \text{ g/mol}} \times 34.0 \text{ g} = 27.77 \text{ g} \]

Key concepts

molar mass

The molar mass is the mass of one mole of a given substance. To find a compound's molar mass, sum the molar masses of all atoms present in the compound. For propane (\text{C}_{3}H_{8}), we have three carbon atoms and eight hydrogen atoms. The molar mass of carbon is 12.01 g/mol, and the molar mass of hydrogen is 1.01 g/mol. So, for propane:
\[ \text{Molar mass of } \text{C}_{3}H_{8} = (3 \times 12.01 \text{ g/mol}) + (8 \times 1.01 \text{ g/mol}) = 36.03 \text{ g/mol} + 8.08 \text{ g/mol} = 44.11 \text{ g/mol} \] This calculation helps us understand the weight of a mole of propane, which we'll use in further calculations.

stoichiometry

Stoichiometry is the field of chemistry that deals with the relative quantities of reactants and products in chemical reactions. It allows you to calculate the required amounts of reactants for a desired amount of product and vice versa. In our propane example, stoichiometry allows us to relate the moles of propane to its mass and the mass of its constituent elements. Understanding stoichiometry is crucial for proper measurements in chemical reactions and ensures that the reactions proceed efficiently.

mole-to-mass conversion

Mole-to-mass conversion involves determining the mass of a given number of moles of a substance. We use the molar mass as a conversion factor. For example, to find the mass of 1.50 moles of propane (\text{C}_{3}H_{8}), we multiply the number of moles by the molar mass of propane:

\[ \text{Mass} = \text{moles} \times \text{molar mass} = 1.50 \text{ moles} \times 44.11 \text{ g/mol} = 66.165 \text{ g} \] This tells us that 1.50 moles of propane weigh 66.165 grams. This type of calculation is commonly used in chemistry when preparing samples for reactions.

mass-to-mole conversion

Mass-to-mole conversion lets us determine the number of moles in a given mass of a substance. The mass is divided by the molar mass. For example, if we have 34.0 grams of propane, we calculate the number of moles as follows:

\[ \text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{34.0 \text{ g}}{44.11 \text{ g/mol}} = 0.771 \text{ moles} \] This shows that 34.0 grams of propane contains approximately 0.771 moles. This calculation is useful for determining the amount of a substance present in a sample.

carbon content in hydrocarbons

Hydrocarbons are compounds composed of carbon and hydrogen atoms. To find the carbon content in a hydrocarbon, we first determine the contribution of carbon atoms to the molar mass. For propane (\text{C}_{3}H_{8}), there are three carbon atoms, each contributing 12.01 grams per mole:

\[ (3 \times 12.01 \text{ g/mol}) = 36.03 \text{ g/mol} \] Then, we calculate the fraction of the molar mass due to carbon and use it to find the mass of carbon in a sample. For 34.0 grams of propane:
\[ \text{Mass of carbon} = \frac{36.03 \text{ g/mol}}{44.11 \text{ g/mol}} \times 34.0 \text{ g} = 27.77 \text{ g} \] This means there are 27.77 grams of carbon in 34.0 grams of propane. Knowing the carbon content is important in applications such as combustion and environmental analysis.