Explanation Given: Mass of gas, m = 0.275 g Volume of gas, V = 158 mL Temperature, T = 25 ∘ C Pressure of gas, P = 556 mmHg Percentage of carbon by mass = 82.66 % Percentage of hydrogen by mass = 17.34 % = Explanation: The percentage by mass of an element gives the number of grams of an element present in the compound. ∴ 82.66 % of carbon by mass represents 82.66 g of carbon. Similarly, 17.34 % of hydrogen represents 17.34 g of hydrogen. The molecular formula of the compound is determined by the following steps: Step 1: Convert the masses of carbon and hydrogen into their moles. The number of moles is given by N u m b e r o f m o l e s = M a s s A t o m i c m a s s The mass of C = 82.66 g and atomic mass of C = 12.0 g/mol ; therefore, moles of C are given by Number of moles of C ( n C ) = 82.66 g 12.0 g /mol = 6.8883 mol Similarly, mass of H = 17.34 g and atomic mass of H = 1.0 g/mol ; therefore, moles of H are given by Number of moles of H ( n H ) = 17.34 g 1.0 g /mol = 17.34 mol Step 2: Convert the moles of carbon and hydrogen into their relative moles by dividing their moles with smallest number of moles. The moles of carbon are small in number; therefore, R e l a t i v e m o l e s o f C = 6.8883 m o l 6.8883 m o l = 1 R e l a t i v e m o l e s o f H = 17.34 m o l 6.8883 m o l = 2.5 Convert these relative moles into whole number by multiplying with 2. Thus, R e l a t i v e m o l e s o f C = 2 R e l a t i v e m o l e s o f H = 5 There are 2 atoms of C and 5 atoms of H ; hence, the empirical formula of the compound is C 2 H 5 . Step 3: Calculate the empirical formula mass. E m p i r i c a l f o r m u l a m a s s = 2 × A t o m i c m a s s o f C + 5 × A t o m i c m a s s o f H = 2 × 12.0 g / m o l + 5 × 1 g /mol = 29

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Chapter 5, Problem 99E

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Chapter 5, Problem 98E

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Masteringchemistry With Pearson Etext -- Valuepack Access Card -- For Principles Of Chemistry: A Molecular Approach

Ch. 5 - Prob. 1SAQ Ch. 5 - Prob. 2SAQ Ch. 5 - Prob. 3SAQ Ch. 5 - Prob. 4SAQ Ch. 5 - Prob. 5SAQ Ch. 5 - Prob. 6SAQ Ch. 5 - Prob. 7SAQ Ch. 5 - Prob. 8SAQ Ch. 5 - Q9. A gas sample at STP contains 1.15 g oxygen and...Ch. 5 - Prob. 10SAQ

Ch. 5 - Prob. 11SAQ Ch. 5 - Prob. 12SAQ Ch. 5 - Prob. 13SAQ Ch. 5 - Prob. 14SAQ Ch. 5 - Prob. 15SAQ Ch. 5 - Prob. 1E Ch. 5 - Prob. 2E Ch. 5 - Prob. 3E Ch. 5 - Prob. 4E Ch. 5 - Prob. 5E Ch. 5 - Prob. 6E Ch. 5 - Prob. 7E Ch. 5 - Prob. 8E Ch. 5 - Prob. 9E Ch. 5 - Prob. 10E Ch. 5 - Prob. 11E Ch. 5 - Prob. 12E Ch. 5 - Prob. 13E Ch. 5 - Prob. 14E Ch. 5 - Prob. 15E Ch. 5 - Prob. 16E Ch. 5 - Prob. 17E Ch. 5 - Prob. 18E Ch. 5 - Prob. 19E Ch. 5 - Prob. 20E Ch. 5 - Prob. 21E Ch. 5 - Prob. 22E Ch. 5 - Prob. 23E Ch. 5 - Prob. 24E Ch. 5 - Prob. 25E Ch. 5 - Prob. 26E Ch. 5 - Prob. 27E Ch. 5 - Prob. 28E Ch. 5 - 29. Given a barometric pressure of 762.4 mmHg,...Ch. 5 - Prob. 30E Ch. 5 - Prob. 31E Ch. 5 - Prob. 32E Ch. 5 - Prob. 33E Ch. 5 - Prob. 34E Ch. 5 - Prob. 35E Ch. 5 - Prob. 36E Ch. 5 - Prob. 37E Ch. 5 - Prob. 38E Ch. 5 - Prob. 39E Ch. 5 - Prob. 40E Ch. 5 - Prob. 41E Ch. 5 - Prob. 42E Ch. 5 - Prob. 43E Ch. 5 - Prob. 44E Ch. 5 - Prob. 45E Ch. 5 - Prob. 46E Ch. 5 - Prob. 47E Ch. 5 - Prob. 48E Ch. 5 - Prob. 49E Ch. 5 - Prob. 50E Ch. 5 - Prob. 51E Ch. 5 - Prob. 52E Ch. 5 - Prob. 53E Ch. 5 - Prob. 54E Ch. 5 - Prob. 55E Ch. 5 - Prob. 56E Ch. 5 - Prob. 57E Ch. 5 - Prob. 58E Ch. 5 - Prob. 59E Ch. 5 - Prob. 60E Ch. 5 - Prob. 61E Ch. 5 - Prob. 62E Ch. 5 - Prob. 63E Ch. 5 - 64. A 275-mL flask contains pure helium at a...Ch. 5 - Prob. 65E Ch. 5 - Prob. 66E Ch. 5 - Prob. 67E Ch. 5 - Prob. 68E Ch. 5 - Prob. 69E Ch. 5 - Prob. 70E Ch. 5 - Prob. 71E Ch. 5 - Prob. 72E Ch. 5 - Prob. 73E Ch. 5 - Prob. 74E Ch. 5 - Prob. 75E Ch. 5 - Prob. 76E Ch. 5 - Prob. 77E Ch. 5 - 78. Ozone is depleted in the stratosphere by...Ch. 5 - Prob. 79E Ch. 5 - Prob. 80E Ch. 5 - Prob. 81E Ch. 5 - Prob. 82E Ch. 5 - Prob. 83E Ch. 5 - Prob. 84E Ch. 5 - Prob. 85E Ch. 5 - Prob. 86E Ch. 5 - Prob. 87E Ch. 5 - Prob. 88E Ch. 5 - Prob. 89E Ch. 5 - Prob. 90E Ch. 5 - Prob. 91E Ch. 5 - Prob. 92E Ch. 5 - Prob. 93E Ch. 5 - Prob. 94E Ch. 5 - 95. Modern pennies are composed of zinc coated...Ch. 5 - Prob. 96E Ch. 5 - Prob. 97E Ch. 5 - Prob. 98E Ch. 5 - Prob. 99E Ch. 5 - Prob. 100E Ch. 5 - Prob. 101E Ch. 5 - Prob. 102E Ch. 5 - Prob. 103E Ch. 5 - Prob. 104E Ch. 5 - Prob. 105E Ch. 5 - Prob. 106E Ch. 5 - Prob. 107E Ch. 5 - Prob. 108E Ch. 5 - Prob. 109E Ch. 5 - Prob. 110E Ch. 5 - Prob. 111E Ch. 5 - Prob. 112E Ch. 5 - Prob. 113E Ch. 5 - Prob. 114E Ch. 5 - Prob. 115E Ch. 5 - Prob. 116E Ch. 5 - Prob. 117E Ch. 5 - Prob. 118E Ch. 5 - Prob. 119E Ch. 5 - Prob. 120E Ch. 5 - Prob. 121E Ch. 5 - Prob. 122E Ch. 5 - Prob. 123E Ch. 5 - Prob. 124E Ch. 5 - Prob. 125E Ch. 5 - Prob. 126E Ch. 5 - Prob. 127E Ch. 5 - Prob. 128E Ch. 5 - Prob. 129E Ch. 5 - Prob. 130E Ch. 5 - Prob. 131E Ch. 5 - Prob. 132E Ch. 5 - Prob. 133E Ch. 5 - Prob. 134E Ch. 5 - Prob. 135E Ch. 5 - Prob. 136E Ch. 5 - Prob. 137E Ch. 5 - Prob. 138E Ch. 5 - Prob. 139E Ch. 5 - Prob. 140E Ch. 5 - Prob. 141E Ch. 5 - Prob. 142E Ch. 5 - Prob. 143E Ch. 5 - Prob. 144E Ch. 5 - Prob. 145E Ch. 5 - Prob. 146E Ch. 5 - Prob. 147E

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