![]() ![]() Multiply by 3 to get rid of the fractional subscript Once again, divide both numbers by the smallest one to get the mole ratio carbon and hydrogen have in the hydrocarbon Now use the molar masses of carbon and hydrogen to determine how many moles of each you have Since every 100 g of water will contain 11.21 g of hydrogen, your hydrocarbon contained #3.447cancel("g"CO_2) * "27.27 g C"/(100cancel("g"CO_2)) = "0.9400 g C"#ĭo the same for water and hydrogen, but keep in mind that you get 2 hydrogen atoms for every molecule of water. ![]() This means that the mass of carbon in the original hydrocarbon was This means that, for every 100 g of #CO_2#, you get 27.27 g of carbon. If you use the molar mass of carbon and that of carbon dioxide, you can determine the percent composition of carbon in #CO_2#. So, you know that the combustion reaction produces 3.447 g of carbon dioxide. This method uses the percent composition of carbon and of hydrogen in carbon dioxide and water, respectively. I'll show you an alternative approach to getting the number of moles of carbon and of hydrogen the hydrocarbon sample contained. Your compound's empirical formula is #C_3H_7#. ![]()
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