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Chapter 3: Problem 162

A certain metal M forms a bromide containing 53.79 percent Br by mass. What is the chemical formula of the compound?

Short Answer

Expert verified
The bromide formed by the metal M has the empirical formula MBr.

Step by step solution

01

Find the moles of each element

Starting with a hypothetical 100 g sample, the number of moles of bromine (Br) and Metal M in this sample is calculated by dividing the mass (in grams) by the respective atomic mass. \nThe atomic mass of Br is approximately 79.90 g/mol. Therefore, the number of moles of Br in the sample is \(53.79 g / 79.90 g/mol \approx 0.673 mol\). \nAs the atomic mass of M is unknown, designate it as x g/mol. The number of moles of metal M is thus \(46.21 g / x g/mol\).
02

Establish a mole ratio

The empirical formula is found by establishing the ratio of the number of moles of each element. The number of moles of bromine is divided by the smaller number of moles, which in this case is itself, so it gives a ratio of 1. For the number of moles of M, it will also be divided by the number of moles of bromine, so the ratio will be \(46.21g / (x g/mol * 0.673 mol)\).
03

Solve the ratio for M

In a bromide, the charge of Br is -1, so the metal M must have a charge of +1 for the compound to be neutral. This means that there must be an equal amount of metal and bromine atoms in the compound, making the ratio calculated in the previous step equal to 1. From the equation \(46.21g / (x g/mol * 0.673 mol) = 1\), the atomic mass of M can be solved as \[ x = 46.21 g / 0.673 mol \approx 68.67 g/mol \].

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Most popular questions from this chapter

Lysine, an essential amino acid in the human body, contains \(\mathrm{C}, \mathrm{H}, \mathrm{O},\) and \(\mathrm{N} .\) In one experiment, the complete combustion of \(2.175 \mathrm{~g}\) of lysine gave \(3.94 \mathrm{~g}\) \(\mathrm{CO}_{2}\) and \(1.89 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) In a separate experiment, \(1.873 \mathrm{~g}\) of lysine gave \(0.436 \mathrm{~g} \mathrm{NH}_{3}\). (a) Calculate the empirical formula of lysine. (b) The approximate molar mass of lysine is \(150 \mathrm{~g}\). What is the molecular formula of the compound?

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Why must a chemical equation be balanced? What law is obeyed by a balanced chemical equation?

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