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Chapter 3: Problem 138

A binary compound between an unknown element \(\mathrm{E}\) and hydrogen contains \(91.27 \% \mathrm{E}\) and \(8.73 \% \mathrm{H}\) by mass. If the formula of the compound is \(\mathrm{E}_{3} \mathrm{H}_{\mathrm{s}}\), calculate the atomic mass of \(\mathrm{E}\).

Short Answer

Expert verified
The atomic mass of the unknown element E is approximately \(3.49 g/mol\).

Step by step solution

01

Gather given information

Percent composition: - Element E: \(91.27%\) - Hydrogen: \(8.73%\) Formula of the compound: \(E_3H_8\)
02

Convert percent composition into grams

To make the calculation easier, assume that we have 100 grams of this compound. This means that we have: - \(91.27 g\) of element E - \(8.73 g\) of hydrogen
03

Find the number of moles for hydrogen and E

To find the atomic mass of element E, first, find the number of moles for both hydrogen and E. The atomic mass of hydrogen is 1 g/mol. So, Hydrogen moles \(\left(n_{H}\right)\) = \(\frac{mass}{molar\ mass}\) \(n_{H}\) = \(\frac{8.73 g}{1 g/mol}\) = \(8.73\) moles Let 'M' represent the atomic mass of element E. Then, E moles \(\left(n_{E}\right)\) = \(\frac{91.27 g}{M g/mol}\)
04

Use the compound formula to establish the mole ratio

The formula \(E_3H_8\) indicates that for every three moles of E, there are eight moles of hydrogen. Using this ratio, we can set up an equation: \(\frac{n_{E}}{n_{H}} = \frac{3}{8}\) Substituting the values: \(\frac{\frac{91.27 g}{M \ g/mol}}{8.73 \ mol}= \frac{3}{8}\)
05

Solve for atomic mass (M)

Solve the equation for the atomic mass 'M': \( \frac{91.27\ g}{M\ g/mol} = \frac{3}{8} \times 8.73\ mol\) Multiply both sides by M: \(91.27\ g = 3 \times 8.73\ mol \times M\) Now, divide both sides by 3 times 8.73: \(M = \frac{91.27\ g}{3 \times 8.73\ mol}\) Calculate M: \(M = 3.49\ g/mol\) #Conclusion# The atomic mass of the unknown element E is approximately \(3.49 g/mol\).

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Most popular questions from this chapter

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