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Chapter 3: Problem 49

A compound whose empirical formula is \(\mathrm{XF}_{3}\) consists of \(65 \%\) \(\mathrm{F}\) by mass. What is the atomic mass of \(\mathrm{X} ?\)

Short Answer

Expert verified
The atomic mass of element X in the compound XF3, which is 65% fluorine by mass, is approximately \(30.69 g/mol\).

Step by step solution

01

Determine the atomic mass of fluorine and calculate the mass of 3 fluorine atoms in the compound

The atomic mass of fluorine (F) is approximately 19 g/mol. Since there are 3 fluorine atoms in the formula XF3, we can calculate the mass of fluorine in the compound: Mass of 3 F atoms = 3 × atomic mass of F = 3 × 19 g/mol = 57 g/mol
02

Calculate the mass of element X in the compound

Since the compound is 65% fluorine by mass, the remaining percentage must be the mass of element X in the compound. Therefore, the mass percentage of element X in the compound is: 100% - 65% = 35% Now, we can determine the mass of element X in the compound by finding the mass equivalent of 35%. Since the mass of 3 F atoms in the compound equals 57 g/mol (calculated in Step 1), we can set up the equation: (35 g X) / (57 g F) = 35% Solve for the mass of X: 35 g X = 0.35 × 57 g F Mass of X = 19.95 g/mol Since the mass percentage of a single atom of X should be the same as the mass percentage in the compound, we can set up the equation: Mass of 1 X atom / (57 g F + Mass of 1 X atom) = 35%
03

Calculate the atomic mass of element X

Now, we will solve the equation we set up in Step 2 to find the atomic mass of element X: Mass of 1 X atom / (57 g F + Mass of 1 X atom) = 35% Let the atomic mass of X be 'm'. m / (57 + m) = 0.35 Solving for 'm', we get: m = 0.35 × (57 + m) m = 19.95 + 0.35m Now, subtract 0.35m from both sides to get: 0.65m = 19.95 Finally, we can find the atomic mass of X by dividing both sides by 0.65: m = 19.95 / 0.65 m ≈ 30.69 g/mol Thus, the atomic mass of element X is approximately 30.69 g/mol.

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Most popular questions from this chapter

\(\mathrm{NO}_{x}\) is a generic term for the nitrogen oxides, \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) \(\mathrm{NO}_{x}\) gases are air pollutants that react to form smog and acid rain. In order to reduce \(\mathrm{NO}_{x}\) emission from vehicle, catalytic converters are installed in car exhausts to decompose NO and \(\mathrm{NO}_{2}\) respectively into \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}(\mathbf{a})\) Write the balanced chemical equations for the decomposition of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) respectively. (b) If the car produces \(100 \mathrm{~g} \mathrm{NO}_{x}\) a day, with equal mole ratio of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\), how many grams of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) are produced respectively?

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