Answer:
For all these questions, we want to find the empirical and molecular formulae of various compounds given their percent composition and molar mass. The technique used to answer one of the questions can accordingly be applied to all of them.
Approaching the first question, we treat the percentages of each element as the mass of that element in a 100 g compound (as the percentages add up to 100%). So, our 100 g compound comprises 7.74 g H and 92.26 g C.
Next, we convert these mass quantities into moles. Divide the mass of each element by its molar mass:
7.74 g H/1.00794 g/mol = 7.679 mol H
92.26 g C/12.0107 g/mol = 7.681 mol C.
Then, we look for the molar quantity that's the smallest ("smaller," in this case, since there are only two), and we divide all the molar quantities by the smallest one. Here, it's a very close call, but the number of moles of H is slightly smaller than that of C. So, we divide each molar quantity by the number of moles of H:
7.679 mol H/7.679 mol H = 1
7.681 mol C/7.679 mol H ≈ 1 C/H (the value is actually slightly larger than 1, but we can treat it as 1 for our purposes).
The quotients we calculated represent the subscripts of our compound's empirical formula, which should provide the most simplified whole number ratio of the elements. So the empirical formula of our compound is C₁H₁, or just CH.
Here, it just so happens that we obtained whole number quotients. If we end up with a quotient that isn't a whole number (e.g., 1.5), we would multiply all the quotients by a common number that <em>would </em>give us the most simplified whole number ratio (so, if we had gotten 1 and 1.5, we'd multiply both by 2, and the empirical formula would have subscripts 2 and 3).
To find the molecular formula (the actual formula of our compound), we use the molar mass of the compound, 78.1134 g/mol. The molar mass of our "empirical compound," CH, is 13.0186 g/mol. Since our empirical formula represents the most simplified molar ratio of the elements, the molar masses of our "empirical compound" and the actual compound should be multiples of one another. We divide 78.1134 g/mol by 13.0176 g/mol and obtain 6. The subscripts in our molecular formula are equal to the subscripts in our empirical formula multiplied by 6.
Thus, our molecular formula is C₆H₆.
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As mentioned before, all the questions here can be answered following the procedure used to answer the first question above. In any case, I've provided the empirical and molecular formulae for the remaining questions below for your reference.
2. Empirical formula: C₁₃H₁₂O; molecular formula: C₁₃H₁₂O
3. Empirical formula: CH; molecular formula: C₈H₈
4. Empirical formula: C₂HCl; molecular formula: C₆H₃Cl₃
5. Empirical formula: Cl₄K₂Pt; molecular formula: Cl₄K₂Pt
6. Empirical formula: C₂H₄Cl; molecular formula: C₄H₈Cl₂
