Question

chemistry 1 worksheet 7 - 3 percent composition & empirical formulas name semaj cohire period
show your work to receive credit. circle your final answer.
a. calculate the percent composition for the following compounds.

1. cr₂o₃
2. ca₃(po₄)₂

b. calculate the percent by mass of iron in each of the following compounds.

1. iron (iii) oxide fe₂o₃
2. iron (ii) oxide feo

c. determine the empirical formula for each compound.

1. a compound contains 0.0130 mol carbon, 0.0390 mol hydrogen, and 0.0065 mol oxygen.
2. a compound consists of 72.2% magnesium and 27.8% nitrogen by mass.
3. glucose contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.

Explanation:

Step1: Calculate molar mass of \(Cr_2O_3\)

The molar mass of \(Cr\) is approximately \(52.00\ g/mol\) and of \(O\) is approximately \(16.00\ g/mol\). For \(Cr_2O_3\), \(M = 2\times52.00+3\times16.00=104 + 48=152\ g/mol\).

Step2: Calculate percent composition of \(Cr\) in \(Cr_2O_3\)

Percent of \(Cr=\frac{2\times52.00}{152}\times100\%=\frac{104}{152}\times100\%\approx68.42\%\). Percent of \(O=\frac{3\times16.00}{152}\times100\%=\frac{48}{152}\times100\%\approx31.58\%\).

Step3: Calculate molar mass of \(Ca_3(PO_4)_2\)

The molar mass of \(Ca\) is approximately \(40.08\ g/mol\), \(P\) is approximately \(30.97\ g/mol\) and \(O\) is approximately \(16.00\ g/mol\). \(M = 3\times40.08+2\times30.97 + 8\times16.00=120.24+61.94 + 128=310.18\ g/mol\).

Step4: Calculate percent composition of elements in \(Ca_3(PO_4)_2\)

Percent of \(Ca=\frac{3\times40.08}{310.18}\times100\%\approx38.76\%\), percent of \(P=\frac{2\times30.97}{310.18}\times100\%\approx19.97\%\), percent of \(O=\frac{8\times16.00}{310.18}\times100\%\approx41.27\%\).

Step5: Calculate molar mass of \(Fe_2O_3\)

The molar mass of \(Fe\) is approximately \(55.85\ g/mol\) and of \(O\) is approximately \(16.00\ g/mol\). \(M = 2\times55.85+3\times16.00 = 111.7+48=159.7\ g/mol\).

Step6: Calculate percent of iron in \(Fe_2O_3\)

Percent of \(Fe=\frac{2\times55.85}{159.7}\times100\%=\frac{111.7}{159.7}\times100\%\approx69.94\%\).

Step7: Calculate molar mass of \(FeO\)

\(M = 55.85+16.00 = 71.85\ g/mol\).

Step8: Calculate percent of iron in \(FeO\)

Percent of \(Fe=\frac{55.85}{71.85}\times100\%\approx77.73\%\).

Step9: Determine empirical - formula for compound in 5

Divide the number of moles of each element by the smallest number of moles (\(n_{O} = 0.0065\ mol\)).
\(C:\frac{0.0130}{0.0065}=2\), \(H:\frac{0.0390}{0.0065}=6\), \(O:\frac{0.0065}{0.0065}=1\). Empirical formula is \(C_2H_6O\).

Step10: Determine empirical - formula for compound in 6

Assume a \(100 - g\) sample. So, \(m_{Mg}=72.2\ g\) and \(m_{N}=27.8\ g\).
The number of moles of \(Mg\), \(n_{Mg}=\frac{72.2}{24.31}\approx2.97\ mol\), the number of moles of \(N\), \(n_{N}=\frac{27.8}{14.01}\approx1.98\ mol\).
Divide by the smaller number of moles (\(n_{N}\)). \(\frac{n_{Mg}}{n_{N}}\approx\frac{2.97}{1.98}\approx1.5\). Multiply by 2 to get whole - numbers. Empirical formula is \(Mg_3N_2\).

Step11: Determine empirical - formula for glucose in 7

Assume a \(100 - g\) sample. \(m_{C}=40.0\ g\), \(m_{H}=6.7\ g\), \(m_{O}=53.3\ g\).
\(n_{C}=\frac{40.0}{12.01}\approx3.33\ mol\), \(n_{H}=\frac{6.7}{1.01}\approx6.63\ mol\), \(n_{O}=\frac{53.3}{16.00}\approx3.33\ mol\).
The empirical formula is \(CH_2O\).

Answer:

1. Percent composition of \(Cr_2O_3\): \(Cr\approx68.42\%\), \(O\approx31.58\%\)
2. Percent composition of \(Ca_3(PO_4)_2\): \(Ca\approx38.76\%\), \(P\approx19.97\%\), \(O\approx41.27\%\)
3. Percent of iron in \(Fe_2O_3\approx69.94\%\)
4. Percent of iron in \(FeO\approx77.73\%\)
5. Empirical formula: \(C_2H_6O\)
6. Empirical formula: \(Mg_3N_2\)
7. Empirical formula: \(CH_2O\)