Combustion analysis of 63.8 mg of a c h and o

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Hey everyone, we're asked to determine the empirical formula of a meal acetate if it is composed of carbon, hydrogen and oxygen. And they've given us the mass of each compound. So we want to take the mass of our compounds and convert them into grams of our atoms. Starting off with carbon, we can start with 0.779g of carbon dioxide. We can convert this into moles of carbon dioxide By using carbon dioxide Molar mass, which is 44.01 g per one mole. And we know that one mole of carbon dioxide Contains one mole of carbon. Since we have only one carbon in our carbon dioxide. And looking at our periodic table, We know one mole of carbon has 12 g. So when we calculate this out, we end up with 0.212 g of carbon. Moving on to Hydrogen, we can take our 0.3254 g of water. And using water smaller mass which is 18.01 g of water Per one mole of water. And we know that one mole of water contains two more of hydrogen and looking at hydrogen Atomic mass, we know that one mole contains 1.01 g of hydrogen And this will take us to a value of 0.0361 g of hydrogen. And lastly to get the grams of oxygen, we can go ahead and take our total amount of ice a mill acetate, which is 0.329g. And we can simply subtract the values that we just calculated. So 0.212g of carbon plus 0.361 g of hydrogen, Which will get us to a value of 0.0809 g of oxygen. Now that we have our grams, we can go ahead and convert into moles. So starting off with 0.212g of carbon, We know that we have 12.01 g of carbon per one mole. When we calculate this out, we end up with 0.0175 mole of carbon. Now to calculate the mole of hydrogen, We start off with our 0.0361 g of hydrogen And we can convert this into a mole of hydrogen by using its atomic mass of 1.01 g per one mole. And this will get us to a value of 0.0357 mole of hydrogen. Moving on to our oxygen, we have 0.0809 g of oxygen and we know that we have 16.0 g of oxygen per one mole of oxygen. This will get us a value of 0. mole of oxygen. Now, in order to get our empirical formula, we're going to have to divide each of our values by the least amount of moles we have here and in this case it's going to be our mall of oxygen, which is 0.5056. So when we divide each of our values, we end up with about 3.5 of carbon seven of hydrogen and one of oxygen. And since we have this decimal point on our carbon, we can't round up so we need to multiply each of our values by two in order to get a whole number. So we end up with seven carbon, 14 hydrogen and two oxygen. So our empirical formula is going to be C seven H 14 02 and this is our final answer. So I hope this made sense and let us know if you have any questions.

Combustion analysis of 63.8 mg of a c h and o

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