![]() ![]() These are not neat and tidy numbers, but they look close to #3/4, 4 1/2 and 1 1/2#. Now, if we add these we have #13.72#mg#, but we know the sample was #35.0#mg#, so the remainder must be #N#: #21.28#mg#. Combustion of 2.90 g of this compound produced 5.79 g of CO2 and 2.37 g of H2O. We can now find the mass of #C# and the mass of #H# in the reactant: An unknown compound contains only C, H, and O. Now each mole of #H# in the reactant produces only #1/2# mole of #H_2O#, so we need to double this number to find #y#, the number of moles of #H# in the reactant: #y=4.6xx10^-3#mol# ![]() This is simply the value of #x#, the number of moles of #C# in the reactant. (note that we had mg of product and grams for molar mass) We know the molar mass of #CO_2# is #44#gmol^-1# and the molar mass of #H_2O# is #18#gmol^-1#.įind the number of moles of #CO_2# in the products: ![]() (There are several oxides of nitrogen such as #N_2O#, but this one corresponds to complete combustion.) The same kind of reasoning applies for #H# to #H_2O# and #N# to #NO_2#. That might look complicated, but all we're saying is that each mole of #C# in the reactant will create #1# mole of #CO_2# as a product, and require #1/2# mole of #O_2# to do so. We can write an equation for the (complete) combustion of the substance as follows:
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