Introduction: Determining Mass of Elements in a Compound

About: Chemistry and electronics have been a staple in my life since I was 8 and have pretty much been my only hobbies although I have dabbled in herbalism, art, music, and various other areas as well.

Warning: Math!

Sometimes someone just needs to know how much of a compound it will to take to produce another substance or just to know how much of an element is in a compound . For example if you have 1kg of CuO waste and want to reclaim the copper from it, it would be helpful to know that CuO is 79.88645477928328% Copper and 20.113545220716722% oxygen by weight which amounts to 798.8645477928328 g copper and 201.13545220716722 g oxygen/ kg CuO. Now obviously only processing 1kg of material wouldn't warrant such a crazy decimal so rounding is pretty handy here which would bring it to a much more manageable and practical 798.9g copper and 201.1g oxygen per kilogram of CuO. I will also be doing an example with the combustion of sugar as well to show a little better of how it can be used in chemical reactions.

Step 1: Find Your Atomic Masses

According to wikipedia, an atomic mass unit (unified) is equal to 1/12 of a carbon 12 atom and is denoted with the letter "u" which stands for units. Incidentally, 1u can also be written as 1 g/mol in the same way that 1 g/ml can be written as 1 kg/L, they are obviously completely different but the numbers are the same. For example in the element oxygen, the atomic mass is 15.9994u so that would also mean that 1 mol of oxygen weighs 15.9994g. In 1mole of a substance there is a set number of atoms or molecules (6.022140857×10^23). Now, in 1 mol CuO there is 1 mol Copper and 1 mol Oxygen, in 1 mol of sugar (C6H12O6) there are 6 mol of carbon, 12 mol of hydrogen, and 6 mol of oxygen. See a pattern yet? Haha

for more information on moles-

One important thing to remember is atomic mass is an average. For a better explanation refer to this page-

Step 2: Do Some Math

First we need to calculate the total atomic mass of our compounds, copper (II) oxide (CuO) and sugar (C6H12O6).

The atomic masses go as follows-
Cu- 63.546u

O- 15.9994u

C- 12.0107u


And now to dive into the actual math and calculate the molar masses

mass(CuO) = (1(mol) * 63.546 (Cu)) + (1(mol) * 15.9994(O)) or if its less confusing-

mass(CuO) = (1 * 63.546) + (1 * 15.9994)

mass(CuO) = 79.5454u or g/mol

mass(sugar) = (6(mol) * 12.0107(C)) + (12(mol) * 1.00794(H)) + (6(mol) * 15.9994(O)) or

mass(sugar) = (6 * 12.0107) + (12 * 1.00794) + (6 * 15.9994)

mass(sugar) = 180.15588u or g/mol

In the case of the copper oxide we would like to know just how much copper is in the oxide so we can find the percentage by-

percentage(Cu) = ( (1(mol) * 63.546(Cu)) / (79.5454(mass(CuO))) ) * 100

percentage(Cu) = ((1 * 63.546) / 79,5454) * 100

percentage(Cu) = 0.7988645477928328 * 100

percentage(Cu) = 79.88645477928328

This means that the total amount of copper in CuO is 79.88645477928328% by weight, and then from that we can find the total mass of the copper in 1kg of CuO-

total(Cu) = 0.7988645477928328 * 1000

total(Cu) = 798.8645477928328g

Step 3: More Math!

Banging your head against the table yet? Haha it gets worse.

Now we are going to calculate how many grams of oxygen it will take to fully combust 1g of sugar and how many grams of CO2 and water are produced as a result.

The first thing we need to do is write out and balance the chemical reaction-

C6H12O6 + 12O → 6CO2 + 6H2O

For more on how to do this, check out this excellent instructable from CrystalE10-

total mass of reactant(O) per mol sugar = 12(mol) * 15.9994

total mass of reactant(O) per mol sugar = 191.9928g

total mass of product(CO2) per mol sugar = 6 * (12.0107(C) + (2 * 15.9994(O)))

total mass of product(CO2) per mol sugar = 264.057g

total mass of product(H2O) per mol sugar = 6 * ((2 * 1.00794) + 15.9994)

total mass of product(H2O) per mol sugar = 108.09168g

mol/g(sugar) =1 / 180.15588(mass(sugar))

mol/g(sugar) = 0.0055507486072616675

total mass of reactant(O) per gram sugar = 191.9928g (total mass of reactant(O) per mol sugar ) * 0.0055507486072616675 (mol/g(sugar)) →

total mass of reactant(O) per gram sugar = 191.9928g * 0.0055507486072616675

total mass of reactant(O) per gram sugar = 1.0657037672042677g

Repeat this for the CO2 and H2O to find that the masses of the reactants are-

sugar - 1g

Oxygen - 1.0657037672042677g

and the masses of the products are-

CO2 - 1.4657140249876943g

H2O - 0.5999897422165739g

and thats it!

So, we found out through a little (mind numbing haha) math that it takes a little over a gram of oxygen to burn a gram of sugar with the products being about 1.5g of CO2 and about 0.6g water.

The great thing about this technique is that it can be applied to any chemical reaction or any set of chemical reactions.

If you want a little more practice doing this you could try working out how much carbon it would take to carbothermally reduce that copper (II) oxide to copper metal and how much CO2 will be produced as a result.

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