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Chapter 9: Problem 33

A compound is found to have the following elemental mass percents: \(\mathrm{Cl}=89.09 \%\), \(\mathrm{C}=10.06 \%, \mathrm{H}=0.84 \% .\) The molar mass of the compound is \(119.378 \mathrm{~g} / \mathrm{mol} .\) What are the empirical and molecular formulas?

Short Answer

Expert verified
The empirical and molecular formulas of the compound are both C\(H_3\)Cl\(3\).

Step by step solution

01

Convert mass percentages to grams

As we're given mass percentages, let's assume we have a 100 g sample of the compound. In this case, the grams of each element are the given percentages: Cl: \(89.09 \ g\) C: \(10.06 \ g\) H: \(0.84 \ g\)
02

Convert grams to moles

Now we need to convert the grams for each element to moles, using their respective molar masses: Molar masses: Cl = \(35.45 \frac{g}{mol}\), C = \(12.01 \frac{g}{mol}\), H = \(1.008 \frac{g}{mol}\) Moles of Cl: \(\frac{89.09 \ g}{35.45 \frac{g}{mol}} = 2.514 \ mol\) Moles of C: \(\frac{10.06 \ g}{12.01 \frac{g}{mol}} = 0.838 \ mol\) Moles of H: \(\frac{0.84 \ g}{1.008 \frac{g}{mol}} = 0.834 \ mol\)
03

Find the mole ratio

To find the mole ratio of each element, we'll divide each mole value by the smallest value: Mole ratio Cl: \(\frac{2.514 \ mol}{0.834 \ mol} = 3.014 \approx 3\) Mole ratio C: \(\frac{0.838 \ mol}{0.834 \ mol} = 1.005 \approx 1\) Mole ratio H: \(\frac{0.834 \ mol}{0.834 \ mol} = 1 \)
04

Determine the empirical formula

Based on the rounded mole ratios, the empirical formula of the compound is: C\(H_3\)Cl\(3\)
05

Calculate the molecular formula

To find the molecular formula, we'll first calculate the empirical formula mass: Empirical formula mass: \(C + 3H + 3Cl = 12.01 + 3(1.008) + 3(35.45) = 119.37 \frac{g}{mol}\) Now we'll divide the molar mass of the compound by the empirical formula mass: \(\frac{119.378 \frac{g}{mol}}{119.37 \frac{g}{mol}} = 1.00067 \approx 1\) Since the ratio is approximately 1, the empirical and molecular formulas are the same: Molecular Formula: C\(H_3\)Cl\(3\)

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Most popular questions from this chapter

Nitrogen and fluorine react to form nitrogen trifluoride according to the balanced chemical equation \(\mathrm{N}_{2}(g)+3 \mathrm{~F}_{2}(g) \rightarrow 2 \mathrm{NF}_{3}(g)\) For each of the following reaction mixtures, choose the limiting reactant: (a) \(0.50 \mathrm{~mol} \mathrm{~N}_{2}(g)\) and \(0.50 \mathrm{~mol} \mathrm{~F}_{2}(g)\) (b) \(12.0 \mathrm{~mol} \mathrm{~N}_{2}(g)\) and \(20.0 \mathrm{~mol} \mathrm{~F}_{2}(g)\) (c) \(2.5 \mathrm{~mol} \mathrm{~N}_{2}(g)\) and \(7.5 \mathrm{~mol} \mathrm{~F}_{2}(g)\) (d) 100 molecules \(\mathrm{N}_{2}(g)\) and 500 molecules \(\mathrm{F}_{2}(g)\) (e) \(5.00 \mathrm{~g} \mathrm{~N}_{2}(g)\) and \(15.0 \mathrm{~g} \mathrm{~F}_{2}(g)\) (f) \(20.0 \mathrm{mg} \mathrm{N}_{2}(g)\) and \(70.0 \mathrm{mg} \mathrm{F}_{2}(g)\)

Which has the greatest mass: 1 mole of ethylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right), 1\) mole of carbon monoxide gas, or 1 mole of nitrogen gas \(\left(\mathrm{N}_{2}\right) ?\)

Nicotine has the formula \(\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{~N}_{2}\). Determine the mass percent of each element.

Sodium (Na) reacts with hydrogen \(\left(\mathrm{H}_{2}\right)\) to form sodium hydride (NaH). A reaction mixture contains \(10.00 \mathrm{~g} \mathrm{Na}\) and \(0.0235 \mathrm{~g} \mathrm{H}_{2}\). (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, \(0.428 \mathrm{~g}\) of \(\mathrm{NaH}\) is recovered. What is the percent yield of the reaction?

How many molecules of aspirin, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\), would be in a tablet that contained \(250 \mathrm{mg}\) of aspirin? How many atoms of carbon would be in the aspirin in that tablet?
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