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Propane is widely used in liquid form as a fuel for barbecue grills and camp stoves. For \(85.5 \mathrm{~g}\) of propane, calculate (a) moles of compound; (b) grams of carbon.

Short Answer

Expert verified
a) 1.94 moles of propane; b) 69.86 grams of carbon.

Step by step solution

01

- Calculate the Moles of Propane (C₃H₈)

First, find the molar mass of propane (C₃H₈). The molar mass can be calculated by summing the atomic masses of all the atoms in the molecule. Carbon (C) has an atomic mass of approximately 12.01 g/mol, and hydrogen (H) has an atomic mass of approximately 1.008 g/mol. \[ \text{Molar Mass of } \text{C}_3\text{H}_8 = (3 \times 12.01) + (8 \times 1.008) = 36.03 + 8.064 = 44.094 \text{ g/mol} \] Next, use the molar mass to find the number of moles by dividing the given mass of propane by the molar mass. \[ \text{Moles of } \text{C}_3\text{H}_8 = \frac{85.5 \text{ g}}{44.094 \text{ g/mol}} = 1.94 \text{ moles} \]
02

- Calculate the Grams of Carbon

Determine the number of moles of carbon in the moles of propane using the molecular formula. Each mole of propane (C₃H₈) contains 3 moles of carbon. \[ \text{Moles of Carbon} = 1.94 \text{ moles of } \text{C}_3\text{H}_8 \times 3 \text{ moles of C per mole of } \text{C}_3\text{H}_8 = 5.82 \text{ moles of C} \] Multiply the moles of carbon by the atomic mass of carbon to get the mass of carbon in grams. \[ \text{Grams of Carbon} = 5.82 \text{ moles of C} \times 12.01 \text{ g/mol} = 69.86 \text{ g} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass
To understand the molar mass, think of it as the weight of one mole of a substance. A mole is a standard unit which equals to Avogadro's number, \(6.022 \times 10^{23}\) particles (atoms, molecules). The molar mass is calculated by adding up the atomic masses of all the atoms in a molecule. For instance, the molecule \( \text{C}_3\text{H}_8 \) contains 3 carbon atoms and 8 hydrogen atoms. Thus, we add the atomic masses of carbon and hydrogen together: \( 3 \times 12.01 \text{ (for C)} + 8 \times 1.008 \text{ (for H)} = 44.094 \text{ g/mol} \).
This value, 44.094 g/mol, is the molar mass of propane. You can find the molar mass of any compound by following these simple steps.
Propane Combustion
Propane, \(C_3H_8 \), is a common fuel used in grills and camping stoves. When propane combusts, it reacts with oxygen to produce carbon dioxide and water. The balanced chemical equation for the combustion of propane is: \[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \].
This equation tells us that one mole of propane reacts with five moles of oxygen to produce three moles of carbon dioxide and four moles of water. This reaction releases energy as heat and is what makes propane a useful fuel.
Atomic Mass
Atomic mass is the weight of a single atom, usually measured in atomic mass units (amu). For example, carbon has an atomic mass of approximately 12.01 amu, while hydrogen is around 1.008 amu. These values are essential for calculating the molar mass of molecules.
When you sum the atomic masses of all atoms in a molecule, you get the molar mass of that molecule. For example, in the molecule \(C_3H_8 \), the atomic masses of the carbon and hydrogen atoms are summed to find the molar mass as seen earlier. This fundamental concept of atomic mass enables chemists to convert between grams and moles, which are crucial for stoichiometric calculations.
By understanding atomic masses, you can grasp how much of a substance is involved in reactions and how different elements contribute to the overall mass of compounds.

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Most popular questions from this chapter

Serotonin \((\mathscr{M}=176 \mathrm{~g} / \mathrm{mol})\) transmits nerve impulses between neurons. It contains \(68.2 \%\) C, \(6.86 \%\) H, \(15.9 \%\) N, and \(9.08 \%\) O by mass. What is its molecular formula?

Chlorine gas can be made in the laboratory by the reaction of hydrochloric acid and manganese(IV) oxide: $$ 4 \mathrm{HCl}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{MnCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2}(g) $$ When 1.82 mol of \(\mathrm{HCl}\) reacts with excess \(\mathrm{MnO}_{2}\), how many (a) moles of \(\mathrm{Cl}_{2}\) and (b) grams of \(\mathrm{Cl}_{2}\) form?

Two successive reactions, \(\mathrm{D} \longrightarrow \mathrm{E}\) and \(\mathrm{E} \longrightarrow \mathrm{F},\) have yields of \(48 \%\) and \(73 \%\), respectively. What is the overall percent yield for conversion of \(\mathrm{D}\) to \(\mathrm{F} ?\)

Is each of the following statements true or false? Correct any that are false. (a) A mole of one substance has the same number of atoms as a mole of any other substance. (b) The theoretical yield for a reaction is based on the balanced chemical equation. (c) A limiting-reactant problem is being stated when the available quantity of one of the reactants is given in moles. (d) The empirical and molecular formulas of a compound are always different.

Which of the following sets of information allows you to obtain the molecular formula of a covalent compound? In each case that allows it, explain how you would proceed (draw a road map and write a plan for a solution). (a) Number of moles of each type of atom in a given sample of the compound (b) Mass \% of each element and the total number of atoms in a molecule of the compound (c) Mass \% of each element and the number of atoms of one element in a molecule of the compound (d) Empirical formula and mass \(\%\) of each element (e) Structural formula

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