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Chapter 9: Problem 150

The beryllium mineral beryl is made up of \(5.03 \%\) by mass Be, \(10.04 \%\) by mass Al, \(31.35 \%\) by mass \(\mathrm{Si}\), and \(53.58 \%\) by mass \(\mathrm{O}\). If the molar mass is approximately \(538 \mathrm{~g} / \mathrm{mol}\), what is the molecular formula for beryl?

Short Answer

Expert verified
The molecular formula for beryl is Be3Al2Si6O18.

Step by step solution

01

Convert the mass percentages to grams

First, assume that there is 100 grams of beryl. This way the mass percentages can be converted directly to grams: - Be: \(5.03 \% * 100 \mathrm{~g} = 5.03 \mathrm{~g}\) - Al: \(10.04 \% * 100 \mathrm{~g} = 10.04 \mathrm{~g}\) - Si: \(31.35 \% * 100 \mathrm{~g} = 31.35 \mathrm{~g}\) - O: \(53.58 \% * 100 \mathrm{~g} = 53.58 \mathrm{~g}\)
02

Determine the moles of each element

Now, we will divide each mass by its respective molar mass to find the moles: - Moles of Be: \(\frac{5.03 \mathrm{~g}}{9.012 \mathrm{~g/mol}} = 0.558 \mathrm{~mol}\) - Moles of Al: \(\frac{10.04 \mathrm{~g}}{26.982 \mathrm{~g/mol}} = 0.372 \mathrm{~mol}\) - Moles of Si: \(\frac{31.35 \mathrm{~g}}{28.085 \mathrm{~g/mol}} = 1.116 \mathrm{~mol}\) - Moles of O: \(\frac{53.58 \mathrm{~g}}{16.00 \mathrm{~g/mol}} = 3.349 \mathrm{~mol}\)
03

Determine the simplest mole ratio

To find the simplest whole-number mole ratio, divide all the mole values by the smallest mole value: - Ratio of Be: \(\frac{0.558 \mathrm{~mol}}{0.372 \mathrm{~mol}} = 1.5\) - Ratio of Al: \(\frac{0.372 \mathrm{~mol}}{0.372 \mathrm{~mol}} = 1\) - Ratio of Si: \(\frac{1.116 \mathrm{~mol}}{0.372 \mathrm{~mol}} = 3\) - Ratio of O: \(\frac{3.349 \mathrm{~mol}}{0.372 \mathrm{~mol}} = 9\) However, the ratio of Be is not a whole number. In order to convert all ratios to whole numbers, multiply everything by 2: - Ratio of Be: \(1.5 * 2 = 3\) - Ratio of Al: \(1 * 2 = 2\) - Ratio of Si: \(3 * 2 = 6\) - Ratio of O: \(9 * 2 = 18\)
04

Write the molecular formula

Now that we have the whole number mole ratios for each element, we can write the molecular formula for beryl: Be3Al2Si6O18 So, the molecular formula for beryl is Be3Al2Si6O18.

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Most popular questions from this chapter

A compound used as an insecticide that contains only \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{Cl}\) is subjected to combustion analysis, yielding \(55.55 \% \mathrm{C}\) and \(3.15 \% \mathrm{H}\). (a) What is the empirical formula of this compound? (b) What is the molecular formula if its empirical formula is \(1 / 3\) the mass of its actual formula?

Ethylene glycol, used for antifreeze, has the molecular formula \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\). Calculate the mass percent of each element in ethylene glycol.

Consider the balanced chemical equation $$ \begin{aligned} &\mathrm{Fe}(\mathrm{CO})_{5}(s)+2 \mathrm{PF}_{3}(l)+\mathrm{H}_{2}(g) \rightarrow \\ &\mathrm{Fe}(\mathrm{CO})_{2}\left(\mathrm{PF}_{3}\right)_{2} \mathrm{H}_{2}(s)+3 \mathrm{CO}(g) \end{aligned} $$ (a) How many grams of \(\mathrm{CO}\) could be produced from \(10.0 \mathrm{~g}\) of \(\mathrm{PF}_{3}\), excess \(\mathrm{Fe}(\mathrm{CO})_{5}\), and excess \(\mathrm{H}_{2} ?\) (b) How many grams of \(\mathrm{CO}\) could be produced from \(5.0\) moles of \(\mathrm{Fe}(\mathrm{CO})_{5}, 8.0\) moles of \(\mathrm{PF}_{3}\), and \(6.0\) moles of \(\mathrm{H}_{2}\) ? (c) How many moles of \(\mathrm{CO}\) could be produced from \(25.0 \mathrm{~g}\) of \(\mathrm{Fe}(\mathrm{CO})_{5}, 10.0 \mathrm{~g}\) of \(\mathrm{PF}_{3}\), and excess \(\mathrm{H}_{2} ?\) (d) The density of hydrogen gas at room temperature and atmospheric pressure is \(0.0820 \mathrm{~g} / \mathrm{L}\). When \(5.00 \mathrm{~L}\) of hydrogen gas at room temperature and atmospheric pressure is mixed with excess \(\mathrm{Fe}(\mathrm{CO})_{5}\) and excess \(\mathrm{PF}_{3}\), the mass of \(\mathrm{CO}\) collected is \(13.5 \mathrm{~g}\). What is the theoretical yield (in grams) and the percent yield of \(\mathrm{CO}\) ?

Nicotine has the formula \(\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{~N}_{2}\). Determine the mass percent of each element.

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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