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Unformatted text preview: Chapter 3: Chemical Compounds 82 77 . Answer: 58.0% M in MO Strategy and Explanation: Given the mass percent of one compound, M 2 O, containing one known element, O, and one unknown element, M, calculate the percent by mass of another compound, MO. Choose a convenient sample mass of M 2 O, such as 100.0 g. Find the mass of M and O in the sample, using the given mass percent. Using the molar mass of oxygen as a conversion factor, determine the number of moles of oxygen, then using the formula stoichiometry of M 2 O as a conversion factor determine the number of moles of M. Find the molar mass of M by dividing the mass of M by the moles of M. Use the molar mass of M, and the formula stoichiometry of MO, to determine the mass percent of M in MO. 73.4% M in M 2 O means that 100.0 grams of M 2 O contains 73.4 grams of M. Mass of O = 100.0 g M 2 O – 73.4 g M = 26.6 g O Formula stoichiometry: 1 mol of M 2 O contains 2 mol M and 1 mol O. 26.6gO " 1molO 15.9994gO " 2molM 1molO = 3.33molM Molar mass of M = massof Minsample molof Minsample = 73.4gM 3.33molM = 22.1 g mol Molar mass of MO = 22.1 g + 15.9994 g = 38.07 g/mol %M = mass of M / molMO mass of MO/ molMO " 100% = 22.1 gM 38.07 gMO " 100% = 58.0% M in MO Reasonable Answer Check: It makes sense that the compound with more atoms of M has a higher mass percent of M. The closest element to M’s atomic mass (22.1) is sodium (atomic mass = 22.99). If M is sodium, the two compounds would probably be sodium oxide (Na 2 O) and sodium peroxide (Na 2 O 2 , a compound made up of two Na + ions and one O 2 2– ion. The simple ratio of Na and O atoms in this compound is 1:1). 78. Answer: 48.203% Fe in FeCO 3 , 69.9426% Fe in Fe 2 O 3 , 72.3591% Fe in Fe 3 O 4 Strategy and Explanation: Given formulas of three compounds, determine the percentage of iron in each of them. Calculate the mass of Fe in one mole of compound, while calculating the molar mass of the compound. Divide the calculated mass of the element by the molar mass of the compound and multiply by 100% to get percent. For FeCO 3 : Mass of Fe per mole of FeCO 3 = 55.845 g Fe Molar mass FeCO 3 = (55.845 g) + (12.0107 g) + 3 × (15.9994 g) = 115.856 g/mol FeCO 3 % Fe = mass of Fe/ mol FeCO 3 mass of FeCO 3 / mol FeCO 3 " 100% = 55.845 g Fe 115.856g FeCO 3 " 100% = 48.203% Fe in FeCO 3 For Fe 2 O 3 : Mass of Fe per mole of Fe 2 O 3 = 2 × (55.845) g Fe = 111.690 g Fe Molar mass Fe 2 O 3 = 111.690 g + 3 × (15.9994 g) = 159.688 g/mol Fe 2 O 3 % Fe = mass of Fe/ mol Fe 2 O 3 mass of Fe 2 O 3 / mol Fe 2 O 3 " 100% = 111.694 g Fe 159.688g Fe 2 O 3 " 100% = 69.9426% Fe in Fe 2 O 3 For Fe 3 O 4 Mass of Fe per mole of Fe 3 O 4 = 3 × (55.847) g Fe = 167.535 g Fe Molar mass Fe 3 O 4 = 167.535 g + 4 × (15.9994 g) = 231.533 g/mol Fe 3 O 4 % Fe = mass of Fe/ mol Fe 3 O 4 mass of Fe 3 O 4 / mol Fe 3 O 4 " 100% = 167.535 g Fe 231.533g Fe 3 O 4 " 100% = 72.3591% Fe in Fe 3 O 4 Reasonable Answer Check: The percentage of iron increases as the formula includes more iron and less of other elements. Chapter 3: Chemical Compounds...
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This note was uploaded on 04/20/2008 for the course CHE 131 taught by Professor Kerber during the Spring '08 term at SUNY Stony Brook.
 Spring '08
 Kerber

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