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11 Select the correct answer from each drop-down menu. Determine the molecular formula of the compound. A compound is used as a gasoline additive. It has a molecular weight of 60.10 atomic mass units and an empirical formula of mathrm(C)_(3) mathrm(H)_(8) mathrm(O) . The molecular formula is mathrm(C) Reset Next

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11
Select the correct answer from each drop-down menu.
Determine the molecular formula of the compound.
A compound is used as a gasoline additive. It has a molecular weight of 60.10 atomic mass units and an empirical formula of mathrm(C)_(3) mathrm(H)_(8) mathrm(O) . The molecular formula is mathrm(C) 
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11 Select the correct answer from each drop-down menu. Determine the molecular formula of the compound. A compound is used as a gasoline additive. It has a molecular weight of 60.10 atomic mass units and an empirical formula of mathrm(C)_(3) mathrm(H)_(8) mathrm(O) . The molecular formula is mathrm(C) Reset Next

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JulianElite · Tutor for 8 years

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The molecular formula of the compound is \(C_3H_8O\).

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## Step1:<br />The molecular formula of a compound can be determined from its empirical formula if the molar mass of the compound is known. The empirical formula represents the smallest whole number ratio of the types of atoms in a compound, while the molecular formula consists of the total number of atoms of each element in one molecule of the compound.<br />## Step2:<br />The empirical formula of the compound given is \(C_3H_8O\). To find the molar mass of this empirical formula, we multiply the atomic mass of each element by the number of atoms of that element in the empirical formula. The atomic mass of Carbon (C) is 12.01, Hydrogen (H) is 1.008, and Oxygen (O) is 16. Therefore, the molar mass of \(C_3H_8O\) is \((3 \times 12.01) + (8 \times 1.008) + 16 = 60.08 \, g/mol\).<br />## Step3:<br />The molar mass of the compound is given as 60.10 g/mol. Since this is very close to the molar mass of the empirical formula, we can conclude that the empirical formula is the same as the molecular formula.
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