Questions: One of the inherently satisfying features of chemistry is that chemical materials form and interact in a rational and predictable manner. For example, one can predict a great deal about a particular material's properties by systematically determining the percentage composition of the elements in that compound. This gives us the relative proportions of the elements in the sample. For a molecular made up of elements A, B, and C, the empirical formula AxByCz means that there are two atoms of C for each atom of A and each atom of B. This may not be the actual chemical formula, which might instead be A2B2C4, but it is the "reduced" version of that formula, called the empirical formula. The actual formula is some multiple of the empirical formula. To know the actual formula we need to know both the empirical formula and the molecular mass of the compound. This provides us with the multiplier value in whole units that must be applied to the empirical formula to get the actual formula. A compound is 40.0% C, 6.70% H, and 53.3% O by mass. Assume that we have a 100-g sample of this compound. Part A What are the subscripts in the empirical formula of this compound? Enter the subscripts for C, H, and O, respectively, separated by commas (e.g., 5,6,7). 1,2,1 The empirical formula of the compound is CH2O. Part B The molecular formula mass of this compound is 120 amu. What are the subscripts in the actual molecular formula? Enter the subscripts for C, H, and O, respectively, separated by commas (e.g., 5,6,7).

One of the inherently satisfying features of chemistry is that chemical materials form and interact in a rational and predictable manner. For example, one can predict a great deal about a particular material's properties by systematically determining the percentage composition of the elements in that compound. This gives us the relative proportions of the elements in the sample. For a molecular made up of elements A, B, and C, the empirical formula AxByCz means that there are two atoms of C for each atom of A and each atom of B. This may not be the actual chemical formula, which might instead be A2B2C4, but it is the "reduced" version of that formula, called the empirical formula. The actual formula is some multiple of the empirical formula. To know the actual formula we need to know both the empirical formula and the molecular mass of the compound. This provides us with the multiplier value in whole units that must be applied to the empirical formula to get the actual formula. A compound is 40.0% C, 6.70% H, and 53.3% O by mass. Assume that we have a 100-g sample of this compound. Part A What are the subscripts in the empirical formula of this compound? Enter the subscripts for C, H, and O, respectively, separated by commas (e.g., 5,6,7). 1,2,1 The empirical formula of the compound is CH2O. Part B The molecular formula mass of this compound is 120 amu. What are the subscripts in the actual molecular formula? Enter the subscripts for C, H, and O, respectively, separated by commas (e.g., 5,6,7).
Transcript text: One of the inherently satisfying features of chemistry is that chemical materials form and interact in a rational and predictable manner. For example, one can predict a great deal about a particular material's properties by systematically determining the percentage composition of the elements in that compound. This gives us the relative proportions of the elements in the sample. For a molecular made up of elements A, B, and C, the empirical formula A$_x$B$_y$C$_z$ means that there are two atoms of C for each atom of A and each atom of B. This may not be the actual chemical formula, which might instead be A$_2$B$_2$C$_4$, but it is the "reduced" version of that formula, called the empirical formula. The actual formula is some multiple of the empirical formula. To know the actual formula we need to know both the empirical formula and the molecular mass of the compound. This provides us with the multiplier value in whole units that must be applied to the empirical formula to get the actual formula. A compound is 40.0% C, 6.70% H, and 53.3% O by mass. Assume that we have a 100-g sample of this compound. Part A What are the subscripts in the empirical formula of this compound? Enter the subscripts for C, H, and O, respectively, separated by commas (e.g., 5,6,7). View Available Hint(s) 1,2,1 Previous Answers Correct The empirical formula of the compound is CH$_2$O. Part B The molecular formula mass of this compound is 120 amu. What are the subscripts in the actual molecular formula? Enter the subscripts for C, H, and O, respectively, separated by commas (e.g., 5,6,7). View Available Hint(s) Submit Previous Answers
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Solution

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Solution Steps

Step 1: Determine the Empirical Formula

Given the percentages of each element in the compound, we assume a 100 g sample. This means we have:

  • 40.0 g of C
  • 6.70 g of H
  • 53.3 g of O

Next, we convert these masses to moles:

  • Moles of C: \(\frac{40.0 \text{ g}}{12.01 \text{ g/mol}} = 3.3314 \text{ mol}\)
  • Moles of H: \(\frac{6.70 \text{ g}}{1.008 \text{ g/mol}} = 6.6468 \text{ mol}\)
  • Moles of O: \(\frac{53.3 \text{ g}}{16.00 \text{ g/mol}} = 3.3313 \text{ mol}\)

We then find the simplest whole number ratio by dividing each by the smallest number of moles:

  • Ratio for C: \(\frac{3.3314}{3.3313} \approx 1\)
  • Ratio for H: \(\frac{6.6468}{3.3313} \approx 2\)
  • Ratio for O: \(\frac{3.3313}{3.3313} \approx 1\)

Thus, the empirical formula is CH\(_2\)O.

Step 2: Determine the Molecular Formula

The molecular formula mass is given as 120 amu. First, we calculate the molar mass of the empirical formula CH\(_2\)O:

  • Molar mass of CH\(_2\)O: \(12.01 + 2(1.008) + 16.00 = 30.026 \text{ g/mol}\)

Next, we find the multiplier by dividing the molecular formula mass by the empirical formula mass:

  • Multiplier: \(\frac{120 \text{ amu}}{30.026 \text{ g/mol}} \approx 4\)

Finally, we multiply the subscripts in the empirical formula by this multiplier to get the actual molecular formula:

  • Subscripts for C: \(1 \times 4 = 4\)
  • Subscripts for H: \(2 \times 4 = 8\)
  • Subscripts for O: \(1 \times 4 = 4\)

Final Answer

\(\boxed{4,8,4}\)

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