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Chapter 9: Problem 112

What is the empirical formula of a compound that is \(17.552 \% \mathrm{Na}, 39.696 \% \mathrm{Cr}\), and \(42.752 \% \mathrm{O} ?\)

Short Answer

Expert verified
The empirical formula of the compound is \( \mathrm{Na}_2\mathrm{Cr}_2\mathrm{O}_7 \).

Step by step solution

01

Convert percentage composition to grams

Let's assume we have 100 grams of the compound - this makes it easier to convert the percentage composition directly to grams. - Na: 17.552% of 100 g = 17.552 g - Cr: 39.696% of 100 g = 39.696 g - O: 42.752% of 100 g = 42.752 g
02

Convert grams to moles

Next, we'll convert the grams of each element to moles using their respective molar masses: - Na: Molar mass = 22.99 g/mol - Cr: Molar mass = 51.996 g/mol - O: Molar mass = 16.00 g/mol Moles of each element: - Moles of Na: \( \frac{17.552 \text{ g}}{22.99 \text{ g/mol}} \) = 0.7632 mol - Moles of Cr: \( \frac{39.696 \text{ g}}{51.996 \text{ g/mol}} \) = 0.7632 mol - Moles of O: \( \frac{42.752 \text{ g}}{16.00 \text{ g/mol}} \) = 2.672 mol
03

Calculate mole ratios

Now, we'll find the mole ratios of each element by dividing the moles of each element by the lowest number of moles: - Mole ratio of Na: \( \frac{0.7632}{0.7632} \) = 1 - Mole ratio of Cr: \( \frac{0.7632}{0.7632} \) = 1 - Mole ratio of O: \( \frac{2.672}{0.7632} \) 鈮� 3.5
04

Simplify to the lowest whole numbers

Since we have a ratio of about 3.5 for O, we need to find the lowest whole number ratios. To do this, we'll multiply each mole ratio by a small whole number (e.g., 2) to obtain: - Mole ratio of Na: 1 脳 2 = 2 - Mole ratio of Cr: 1 脳 2 = 2 - Mole ratio of O: 3.5 脳 2 = 7 So, the empirical formula of the compound is: \( \mathrm{Na}_2\mathrm{Cr}_2\mathrm{O}_7 \)

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Most popular questions from this chapter

In a chemical reaction, the mass of the products is equal to the mass of the reactants consumed. Are the moles of product equal to the moles of reactant consumed? Explain your answer.

Consider the following balanced chemical equation: \(2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) (a) How many grams of water are formed from \(5.00 \mathrm{~g}\) of \(\mathrm{H}_{2}\) and an excess amount of \(\mathrm{O}_{2}\) ? (b) How many grams of \(\mathrm{O}_{2}\) do you need to produce \(5.00 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) ? (c) Given \(100.0 \mathrm{~g}\) of \(\mathrm{H}_{2}\), how many grams of \(\mathrm{O}_{2}\) are required to run the reaction in a stoichiometric fashion? (d) What is the theoretical yield in grams of water upon combining \(50.0 \mathrm{~g}\) of \(\mathrm{O}_{2}\) with an excess amount of \(\mathrm{H}_{2}\) ? (e) Express the answer to part (d) in terms of the number of water molecules.

Consider the unbalanced chemical equation \(\mathrm{S}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{SO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (a) Balance the equation. (b) If you react \(4.80 \mathrm{~g}\) of sulfur with \(16.20 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\), how many grams of \(\mathrm{SO}_{2}\) can theoretically be produced?

Consider the unbalanced chemical equation \(\mathrm{CaC}_{2}+\mathrm{CO} \rightarrow \mathrm{C}+\mathrm{CaCO}_{3}\) When the reaction is complete, \(135.4 \mathrm{~g}\) of \(\mathrm{CaCO}_{3}\) produced and \(38.5 \mathrm{~g}\) of \(\mathrm{CaC}_{2}\) is left over. Assuming the reaction had a \(100 \%\) yield, what were the mass of the two reactants at the beginning of the reaction

Determine the mass percent of each element in aluminum sulfate.
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