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Chapter 9: Problem 7

If you have 1 mole of propane, \(\mathrm{C}_{3} \mathrm{H}_{8}\), a gas used in outdoor grills and industrial torches: (a) How many propane molecules do you have? (b) How many hydrogen atoms do you have? (c) What is the mass in grams of 1 mole of propane? Answer: (a) Having 1 mole of propane molecules means you have \(6.022 \times 10^{23}\) propane molecules. (That is the meaning of 1 mole.) (b) You know that 1 mole of propane contains 8 moles of hydrogen atoms. (That is what the subscript 8 in the formula \(\mathrm{C}_{3} \mathrm{H}_{8}\) tells you.) You also know there are \(6.022 \times 10^{23} \mathrm{H}\) atoms per mole of them, which means you have:

Short Answer

Expert verified
(a) In 1 mole of propane, there are \(6.022 \times 10^{23}\) propane molecules. (b) In 1 mole of propane, there are \(4.8176 \times 10^{24}\) hydrogen atoms. (c) The mass of 1 mole of propane is 44.11 grams.

Step by step solution

01

Calculate the number of propane molecules

Since we have 1 mole of propane, we can use Avogadro's number to find the number of propane molecules. Avogadro's number is \(6.022 \times 10^{23}\), which is the number of atoms or molecules in 1 mole of a substance. Therefore, we have: \(1 \: \text{mole} \: \mathrm{C}_3\mathrm{H}_8 \times \frac{6.022 \times 10^{23} \: \text{molecules}}{1 \: \text{mole}} = 6.022 \times 10^{23} \: \text{molecules}\)
02

Calculate the number of hydrogen atoms

We know that each propane molecule has 8 hydrogen atoms, which is indicated by the subscript 8 in the formula \(\mathrm{C}_{3}\mathrm{H}_{8}\). Since we have \(6.022 \times 10^{23}\) propane molecules, we can calculate the total number of hydrogen atoms as follows: \(6.022 \times 10^{23} \: \text{molecules} \times \frac{8 \: \mathrm{H} \: \text{atoms}}{1 \: \mathrm{C}_3\mathrm{H}_8 \: \text{molecule}} = 4.8176 \times 10^{24} \: \mathrm{H} \: \text{atoms}\)
03

Calculate the mass of 1 mole of propane

To find the mass of 1 mole of propane, we need to sum the molar masses of the individual atoms in the chemical formula. The molar mass of carbon (C) is 12.01 g/mol, and the molar mass of hydrogen (H) is 1.01 g/mol. Therefore, the mass of 1 mole of propane is: \((3 \times 12.01 \: \frac{\text{g}}{\text{mol}}) + (8 \times 1.01 \: \frac{\text{g}}{\text{mol}}) = 36.03 + 8.08 = 44.11 \: \frac{\text{g}}{\text{mol}}\) Answer: (a) There are \(6.022 \times 10^{23}\) propane molecules in 1 mole of propane. (b) There are \(4.8176 \times 10^{24}\) hydrogen atoms in 1 mole of propane. (c) The mass of 1 mole of propane is 44.11 grams.

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Most popular questions from this chapter

Consider the unbalanced chemical equation \(\mathrm{CaCN}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CaCO}_{3}+\mathrm{NH}_{3}\) If you began the reaction with \(5.65 \mathrm{~g}\) of \(\mathrm{CaCN}_{2}\) and \(12.2 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\), how many grams of \(\mathrm{NH}_{3}\) would be produced if the reaction had an \(86.0 \%\) yield?

Sodium metal reacts with water to form aqueous sodium hydroxide and hydrogen gas. (a) Write a balanced chemical equation for this reaction. (b) Which reactant is limiting if \(100.0 \mathrm{~g}\) of \(\mathrm{Na}\) and \(4.00\) moles of water are used?

\(5.00 \mathrm{~g}\) of solid sodium (Na) and \(30.0 \mathrm{~g}\) of liquid bromine \(\left(\mathrm{Br}_{2}\right)\) react to form solid \(\mathrm{NaBr}\). (a) Write a balanced chemical equation for this reaction. (b) Which reactant is limiting? (c) What is the theoretical yield for this reaction in grams? (d) How many grams of excess reactant are left over at the end of the reaction? (e) When this reaction is actually performed, \(14.7 \mathrm{~g}\) of \(\mathrm{NaBr}\) is recovered. What is the percent yield of the reaction?

Aluminum metal burns in chlorine gas to form aluminum chloride. (a) Write a balanced chemical equation for this reaction. (b) Which reactant is limiting if \(100.0 \mathrm{~g}\) of \(\mathrm{Al}\) and \(5.00\) moles of \(\mathrm{Cl}_{2}\) are used?

What is the maximum possible value of the percent yield of a chemical reaction?
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