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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: \( \frac{2.672}{0.7632} \) ≈ 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 × 2 = 2 - Mole ratio of Cr: 1 × 2 = 2 - Mole ratio of O: 3.5 × 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

Tetraphosphorus decoxide, \(\mathrm{P}_{4} \mathrm{O}_{10}\), reacts with water to form phosphoric acid. If \(52.5 \mathrm{~g}\) of \(\mathrm{P}_{4} \mathrm{O}_{10}\) reacted with \(25.0 \mathrm{~g}\) of water, how many grams of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) could theoretically be produced?

Succinic acid, an intermediate in the metabolism of certain foods, has a molecular mass of \(118.1 \mathrm{~g} / \mathrm{mole}\). A \(1.926 \mathrm{~g}\) sample of succinic acid \(\left(\mathrm{H}_{x} \mathrm{Suc}\right)\) reacts with exactly \(1.25 \mathrm{~g}\) of \(\mathrm{NaOH}\) according to the following balanced equation: \(\mathrm{H}_{x} \mathrm{Suc}+x \mathrm{NaOH} \rightarrow \mathrm{Na}_{x} \mathrm{Suc}+x \mathrm{H}_{2} \mathrm{O}\) What is the value of \(x\) ?

For the reaction \(\mathrm{BF}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{3} \mathrm{BO}_{3}+\mathrm{HBF}_{4}\) (a) Balance the equation. (b) If you react \(24.2 \mathrm{~g}\) of \(\mathrm{BF}_{3}\) with an excess of water and generate \(14.8 \mathrm{~g}\) of \(\mathrm{HBF}_{4}\), what is your percent yield?

Chlorine \(\left(\mathrm{Cl}_{2}\right)\) and fluorine \(\left(\mathrm{F}_{2}\right)\) react to form \(\mathrm{ClF}_{3}\). A reaction vessel contains \(2.50\) moles of \(\mathrm{Cl}_{2}\) and \(6.15\) moles of \(\mathrm{F}_{2}\). (a) Write a balanced chemical equation for this reaction. (b) Which reactant is limiting?

Consider the balanced chemical equation \(2 \mathrm{~A}+\mathrm{B} \rightarrow 2 \mathrm{C}+\mathrm{D}\) When \(8.0 \mathrm{~g}\) of A reacts completely with \(6.0 \mathrm{~g}\) of \(\mathrm{B}\), \(10.0 \mathrm{~g}\) of \(\mathrm{C}\) and \(4.0 \mathrm{~g}\) of \(\mathrm{D}\) are produced. Assuming the yield is \(100 \%\), (a) Which has a greater molar mass, A or C? (b) Which has a greater molar mass, A or B? (c) Which has a greater molar mass, A or D? (d) If the molar mass of \(\mathrm{A}\) is \(24.0 \mathrm{~g} / \mathrm{mol}\), determine the molar mass of \(\mathrm{B}, \mathrm{C}\), and \(\mathrm{D}\).
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