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Chapter 3: Problem 81

A compound containing only sulfur and nitrogen is \(69.6 \% \mathrm{~S}\) by mass; the molar mass is \(184 \mathrm{~g} / \mathrm{mol}\). What are the empirical and molecular formulas of the compound?

Short Answer

Expert verified
The empirical formula of the compound containing only sulfur and nitrogen is SN, with a 1:1 ratio of sulfur and nitrogen. The molecular formula of the compound is S鈧凬鈧�.

Step by step solution

01

Calculate the mass of each element in a 100 g sample

It is easiest to work with 100 g of the compound since the composition is given in percentage. In a 100 g sample, there would be 69.6 g of sulfur and the remaining mass would be nitrogen (100 g - 69.6 g), which is 30.4 g.
02

Convert mass to moles

Use the molar mass of each element to convert the mass of sulfur and nitrogen into moles. The molar mass of sulfur (S) is 32.06 g/mol, and the molar mass of nitrogen (N) is 14.01 g/mol. Moles of S: \( \frac{69.6\,\mathrm{g}}{32.06\,\mathrm{g/mol}} = 2.17\,\mathrm{mol} \) Moles of N: \( \frac{30.4\,\mathrm{g}}{14.01\,\mathrm{g/mol}} = 2.17\,\mathrm{mol} \)
03

Determine the simplest whole-number ratio

To find the simplest whole-number ratio of S and N, divide both the moles of S and moles of N by the smallest number of moles, in this case, 2.17. Simplest ratio of S: \( \frac{2.17\,\mathrm{mol}}{2.17\,\mathrm{mol}} = 1 \) Simplest ratio of N: \( \frac{2.17\,\mathrm{mol}}{2.17\,\mathrm{mol}} = 1 \) The empirical formula thus is SN, with a 1:1 ratio of sulfur and nitrogen.
04

Calculate the molar mass of the empirical formula

To find the molecular formula, we will first determine the molar mass of the empirical formula. Molar mass of SN: 32.06 g/mol (S) + 14.01 g/mol (N) = 46.07 g/mol
05

Determine the molecular formula

To find the molecular formula of the compound, we will use the ratio of the molar mass of the molecular formula (184 g/mol) to the molar mass of the empirical formula (46.07 g/mol). Molecular formula ratio: \( \frac{184\,\mathrm{g/mol}}{46.07\,\mathrm{g/mol}} = 4 \) Multiply the empirical formula by the ratio (4) to find the molecular formula: \( 4 \times \mathrm{SN} = \mathrm{S_4N_4} \) The molecular formula of the compound is S鈧凬鈧�.

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Most popular questions from this chapter

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ What mass of \(\mathrm{NH}_{3}\) must be used to produce \(1.0 \times 10^{6} \mathrm{~kg} \mathrm{HNO}_{3}\) by the Ostwald process? Assume \(100 \%\) yield in each reaction and assume that the NO produced in the third step is not recycled.

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The molecular formula of acetylsalicylic acid (aspirin), one of the most commonly used pain relievers, is \(\mathrm{C}_{9} \mathrm{H}_{3} \mathrm{O}_{4}\). a. Calculate the molar mass of aspirin. b. A typical aspirin tablet contains \(500 . \mathrm{mg} \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4} .\) What amount (moles) of \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\) molecules and what number of molecules of acetylsalicylic acid are in a \(500 .-\mathrm{mg}\) tablet?

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