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Chpt 3 Solutions

# Chpt 3 Solutions - CHAPTER 3 STOICHIOMETRY Questions 23...

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44 CHAPTER 3 STOICHIOMETRY Questions 23. Isotope Mass Abundance 12 C 0 000 . 12 amu 98.89% 13 C 13.034 amu 1.11% Average mass = 0.9889 ) 0 000 . 12 ( + 0.0111(13.034) = 12.01 amu From the relative abundances, there would be 9889 atoms of 12 C and 111 atoms of 13 C in the 10,000 atom sample. The average mass of carbon is independent of the sample size; it will always be 12.01 amu. Total mass = 10,000 atoms × atom amu 01 . 12 = 1.201 × 10 5 amu For 1 mole of carbon (6.0221 × 10 23 atoms C), the average mass would still be 12.01 amu. The number of 12 C atoms would be 0.9889(6.0221 × 10 23 ) = 5.955 × 10 23 atoms 12 C, and the number of 13 C atoms would be 0.0111(6.0221 × 10 23 ) = 6.68 × 10 21 atoms 13 C. Total mass = 6.0221 × 10 23 atoms × atom amu 01 . 12 = 7.233 × 10 24 amu Total mass in g = 6.0221 × 10 23 atoms × atom amu 01 . 12 × amu 10 0221 . 6 g 1 23 × = 12.01 g/mol By using the carbon-12 standard to define the relative masses of all of the isotopes, as well as to define the number of things in a mole, then each element’s average atomic mass in units of grams is the mass of a mole of that element as it is found in nature. 24. Consider a sample of glucose, C 6 H 12 O 6 . The molar mass of glucose is 180.16 g/mol. The chemical formula allows one to convert from molecules of glucose to atoms of carbon, hydrogen, or oxygen present and vice versa. The chemical formula also gives the mole relationship in the formula. One mole of glucose contains 6 mol C, 12 mol H, and 6 mol O. Thus mole conversions between molecules and atoms are possible using the chemical for- mula. The molar mass allows one to convert between mass and moles of compound, and Avogadro’s number (6.022 × 10 23 ) allows one to convert between moles of compound and number of molecules.

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CHAPTER 3 STOICHIOMETRY 45 25. Avogadro’s number of dollars = dollars mol dollars 10 022 . 6 23 × people 10 6 dollars mol dollars 10 022 . 6 dollars mol 1 9 23 × × × = 1 × 10 14 dollars/person 1 trillion = 1,000,000,000,000 = 1 × 10 12 ; each person would have 100 trillion dollars. 26. The molar mass is the mass of 1 mole of the compound. The empirical mass is the mass of 1 mole of the empirical formula. The molar mass is a whole-number multiple of the empirical mass. The masses are the same when the molecular formula = empirical formula, and the masses are different when the two formulas are different. When different, the empirical mass must be multiplied by the same whole number used to convert the empirical formula to the molecular formula. For example, C 6 H 12 O 6 is the molecular formula for glucose, and CH 2 O is the empirical formula. The whole-number multiplier is 6. This same factor of 6 is the mul- tiplier used to equate the empirical mass (30 g/mol) of glucose to the molar mass (180 g/mol). 27. The mass percent of a compound is a constant no matter what amount of substance is present. Compounds always have constant composition. 28. A balanced equation starts with the correct formulas of the reactants and products. The co- efficients necessary to balance the equation give molecule relationships as well as mole relationships between reactants and products. The state (phase) of the reactants and products is also given. Finally, special reaction conditions are sometimes listed above or below the arrow. These can include special catalysts used and/or special temperatures required for a
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Chpt 3 Solutions - CHAPTER 3 STOICHIOMETRY Questions 23...

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