##”C”_8″H”_8″O”_3##

The idea here is that you can use the to figure out how much carbon and hydrogen you had in your ##”2.500 g”## sample of vanillin.

When vanillin undergoes combustion, all the carbon that was present in the sample will now be a part of the carbon dioxide. Similarly, all the hydrogen that was present in the sample will now be a part of the water.

Carbon dioxide, ##”CO”_2##, has a molar mass of ##”44.01 g mol”^(-1)##, which implies that one mole of carbon dioxide has a mass of ##”44.01 g”##.

This means that the ##”5.79 g”## of carbon dioxide will be equivalent to

##5.79 color(red)(cancel(color(black)(“g”))) * “1 mole CO”_2/(44.01color(red)(cancel(color(black)(“g”)))) = “0.13156 moles CO”_2##

As you know, every mole of carbon dioxide contains ##1## mole of carbon. Use carbon’s molar mass to figure out how much carbon you have in this sample

##0.13156 color(red)(cancel(color(black)(“moles CO”_2))) * (1 color(red)(cancel(color(black)(“mole C”))))/(1color(red)(cancel(color(black)(“mole CO”_2)))) * “12.011 g”/(1color(red)(cancel(color(black)(“mole C”)))) = “1.5802 g C”##

Water, ##”H”_2″O”##, has a molar mass of ##”18.015 g mol”^(-1)##. Use this value to calculate how many moles of water were produced by the reaction

##1.18 color(red)(cancel(color(black)(“g”))) * (“1 mole H”_2″O”)/(18.015color(red)(cancel(color(black)(“g”)))) = “0.065501 moles H”_2″O”##

Since every mole of water contains ##2## moles of hydrogen, and since hydrogen has a molar mass of ##”1.00794 g mol”^(-1)##, you can say that the sample contained

##0.065501 color(red)(cancel(color(black)(“moles H”_2″O”))) * (2color(red)(cancel(color(black)(“moles H”))))/(1color(red)(cancel(color(black)(“mole H”_2″O”)))) * “1.00794 g”/(1color(red)(cancel(color(black)(“mole H”)))) = “0.1320 g H”##

You know that vanillin contains carbon, hydrogen, and oxygen, so use the masses of carbon and hydrogen to calculate how many grams of oxygen were present in the original sample

##m_”vanillin” = m_”C” + m_”H” + m_”O”##

##m_”O” = “2.500 g” – “1.5802 g” – “0.1320 g” = “0.7878 g O”##

Now, in order to find the compound’s empirical formula, you must find the smallest whole number ratio that exists between carbon, hydrogen, and oxygen in vanillin.

To do that, convert the masses to moles by using molar masses

##”For C: ” 1.5802 color(red)(cancel(color(black)(“g”))) * “1 mole C”/(12.011color(red)(cancel(color(black)(“g”)))) = “0.13156 moles C”##

##”For H: ” 0.1320 color(red)(cancel(color(black)(“g”))) * “1 mole H”/(1.00794 color(red)(cancel(color(black)(“g”)))) = “0.1310 moles H”##

##”For O: ” 0.7878 color(red)(cancel(color(black)(“g”))) * “1 mole O”/(15.9994 color(red)(cancel(color(black)(“g”)))) = “0.04924 moles O”##

To find the mole ratio that exists between the , divide all values by the smallest one to find

##”For C: ” (0.13156 color(red)(cancel(color(black)(“moles”))))/(0.04924color(red)(cancel(color(black)(“moles”)))) = 2.67##

##”For H: ” (0.1310color(red)(cancel(color(black)(“moles”))))/(0.04924color(red)(cancel(color(black)(“moles”)))) = 2.66##

##”For O: ” (0.04924color(red)(cancel(color(black)(“moles”))))/(0.04924color(red)(cancel(color(black)(“moles”)))) = 1##

Now, you’re looking for the smallest whole number ratio that exists between the three elements, so multiply all values by ##3## to get

##”C”_ ((2.67 * 3)) “H”_ ((2.66 * 3)) “O”_ ((1 * 3)) implies color(green)(|bar(ul(color(white)(a/a)color(black)(“C”_ 8″H”_ 8″O”_3)color(white)(a/a)|)))##

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