chemistry 1 worksheet 7 - 3 percent composition & empirical formulas name semaj cohire period \nshow your…

chemistry 1 worksheet 7 - 3 percent composition & empirical formulas name semaj cohire period \nshow your work to receive credit. circle your final answer.\na. calculate the percent composition for the following compounds.\n1. cr₂o₃\n2. ca₃(po₄)₂\n\nb. calculate the percent by mass of iron in each of the following compounds.\n3. iron (iii) oxide fe₂o₃\n4. iron (ii) oxide feo\n\nc. determine the empirical formula for each compound.\n5. a compound contains 0.0130 mol carbon, 0.0390 mol hydrogen, and 0.0065 mol oxygen.\n6. a compound consists of 72.2% magnesium and 27.8% nitrogen by mass.\n7. glucose contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.

chemistry 1 worksheet 7 - 3 percent composition & empirical formulas name semaj cohire period \nshow your work to receive credit. circle your final answer.\na. calculate the percent composition for the following compounds.\n1. cr₂o₃\n2. ca₃(po₄)₂\n\nb. calculate the percent by mass of iron in each of the following compounds.\n3. iron (iii) oxide fe₂o₃\n4. iron (ii) oxide feo\n\nc. determine the empirical formula for each compound.\n5. a compound contains 0.0130 mol carbon, 0.0390 mol hydrogen, and 0.0065 mol oxygen.\n6. a compound consists of 72.2% magnesium and 27.8% nitrogen by mass.\n7. glucose contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.

Answer

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)