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Chapter 3: Problem 83

A compound that contains only carbon, hydrogen, and oxygen is \(48.64 \% \mathrm{C}\) and \(8.16 \% \mathrm{H}\) by mass. What is the empirical formula of this substance?

Short Answer

Expert verified
The empirical formula for the given compound is \(C_3H_6O_2\).

Step by step solution

01

Convert percentages to grams

Since we are given the percentage composition by mass, let's assume we have a 100 g sample of the substance. This way, the percentages directly correspond to grams: - Carbon: 48.64% of 100 g = 48.64 g C - Hydrogen: 8.16% of 100 g = 8.16 g H - Oxygen: 100% - (48.64% + 8.16%) = 43.20% of 100 g = 43.20 g O
02

Convert grams to moles

Now, we'll convert the grams of each element into moles using their respective molar masses: Carbon: \( \frac{48.64 \, \text{g} }{12.01 \, \frac{\text{g}}{\text{mol}} } = 4.05 \, \text{mol} \) Hydrogen: \( \frac{8.16 \, \text{g} }{1.008 \, \frac{\text{g}}{\text{mol}} } = 8.10 \, \text{mol} \) Oxygen: \( \frac{43.20 \, \text{g} }{16.00 \, \frac{\text{g}}{\text{mol}} } = 2.70 \, \text{mol} \)
03

Find the simplest whole-number ratio

We'll divide each of the calculated values in moles by the smallest value (which is 2.70 mol), and round the result to the nearest whole number: - Carbon: \( \frac{4.05 \, \text{mol}}{2.70 \, \text{mol}} \approx 1.50 \rightarrow 1.5 \approx 1.5 \) - Hydrogen: \( \frac{8.10 \, \text{mol} }{2.70 \, \text{mol}} \approx 3.00 \rightarrow 3.0 \approx 3 \) - Oxygen: \( \frac{2.70 \, \text{mol} }{2.70 \, \text{mol}} = 1.00 \rightarrow 1.0 \approx 1 \) Since the ratio of carbon is approximately 1.5, we'll multiply all the ratios by 2 to get whole numbers: - Carbon: 1.5 × 2 = 3 - Hydrogen: 3 × 2 = 6 - Oxygen: 1 × 2 = 2
04

Write the empirical formula

Based on the whole-number ratios found in step 3, the empirical formula for this compound is: \(C_3H_6O_2\)

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Most popular questions from this chapter

Which of the following statements about chemical equations is(are) true? a. When balancing a chemical equation, you can never change the coefficient in front of any chemical formula. b. The coefficients in a balanced chemical equation refer to the number of grams of reactants and products. c. In a chemical equation, the reactants are on the right and the products are on the left. d. When balancing a chemical equation, you can never change the subscripts of any chemical formula. e. In chemical reactions, matter is neither created nor destroyed so a chemical equation must have the same number of atoms on both sides of the equation.

A \(2.25-g\) sample of scandium metal is reacted with excess hydrochloric acid to produce \(0.1502 \mathrm{~g}\) hydrogen gas. What is the formula of the scandium chloride produced in the reaction?

Many cereals are made with high moisture content so that the cereal can be formed into various shapes before it is dried. \(\mathrm{A}\) cereal product containing \(58 \% \mathrm{H}_{2} \mathrm{O}\) by mass is produced at the rate of \(1000 . \mathrm{kg} / \mathrm{h}\). What mass of water must be evaporated per hour if the final product contains only \(20 . \%\) water?

There are two binary compounds of mercury and oxygen. Heating either of them results in the decomposition of the compound, with oxygen gas escaping into the atmosphere while leaving a residue of pure mercury. Heating \(0.6498 \mathrm{~g}\) of one of the compounds leaves a residue of \(0.6018 \mathrm{~g}\). Heating \(0.4172 \mathrm{~g}\) of the other compound results in a mass loss of \(0.016 \mathrm{~g}\). Determine the empirical formula of each compound.

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?
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