Chapter 3: Problem 31
How is the mass percent of elements in a compound different for a 1.0-g sample versus a 100.-g sample versus a 1-mole sample of the compound?
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Chapter 3: Problem 31
How is the mass percent of elements in a compound different for a 1.0-g sample versus a 100.-g sample versus a 1-mole sample of the compound?
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Elixirs such as Alka-Seltzer use the reaction of sodium bicarbonate with citric acid in aqueous solution to produce a fizz: $$ 3 \mathrm{NaHCO}_{3}(a q)+\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}(a q) \longrightarrow $$ $$ 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) $$ a. What mass of \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\) should be used for every \(1.0 \times 10^{2} \mathrm{mg} \mathrm{NaHCO}_{3} ?\) b. What mass of \(\mathrm{CO}_{2}(g)\) could be produced from such a mixture?
Natural rubidium has the average mass of 85.4678 \(\mathrm{u}\) and is composed of isotopes \(^{85} \mathrm{Rb}(\mathrm{mass}=84.9117 \mathrm{u})\) and \(^{87} \mathrm{Rb}\) . The ratio of atoms \(^{85} \mathrm{Rb} /^{87} \mathrm{Rb}\) in natural rubidium is \(2.591 .\) Calculate the mass of \(^{87} \mathrm{Rb}\) .
Considering your answer to Exercise \(81,\) which type of formula, empirical or molecular, can be obtained from elemental analysis that gives percent composition?
A sample of urea contains \(1.121 \mathrm{g} \mathrm{N}, 0.161 \mathrm{g} \mathrm{H}, 0.480 \mathrm{g} \mathrm{C}\) and 0.640 \(\mathrm{g}\) O. What is the empirical formula of urea?
A \(2.25-\) g sample of scandium metal is reacted with excess hydrochloric acid to produce 0.1502 g hydrogen gas. What is the formula of the scandium chloride produced in the reaction?
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