1. Author: J R Reid The Mole Mass Relative Atomic Mass Amount Calculations

2. Mass Mass is the fancy scientific name for the amount of matter or stuff in an object Mass has the symbol m, and is measured in grams (g) Mass is not the same thing as weight For example: If you have 70kg of mass here on earth and someone shoots you into space where there is no gravity, you do not lose any matter (stuff). However, in space you may weigh nothing at all – but you still have mass. Weight is measured in Newtons – it is the effect of gravity on a mass.

3. Amount ‘Amount’ is the scientific name for the number of particles in an object. The particles could be atoms or molecules Amount has the symbol n, and is measured in moles (mol) One mole is equal to 6x10 23 particles. That is 600,000,000,000,000,000,000,000. This number is called Avogadro’s number

4. Atomic Mass Each element on the periodic table has a different mass. Hydrogen is the smallest, helium is next… The mass is caused by the number of protons and neutrons each element has in its nucleus The mass number is the amount of grams per mole an element has: For example: Uranium can have a mass number of 235 grams per mole. This means that 600,000,000,000,000,000,000,000 Uranium atoms has a mass of 235 grams. This can be simplified by writing it like this: 235gmol -1

5. Relative Atomic Mass Unfortunately when we look up the periodic table we find that the mass numbers are not usually whole numbers. This is because here on Earth each element can have a number of isotopes. An isotope is a different arrangement of the nucleus, for example hydrogen can have a mass number of 1, or 2, or 3. This is because they can have different amounts of neutrons but still be hydrogen The Relative Atomic Mass (A r ) on the periodic table is the mean mass of that element here on Earth For example: Hydrogen has a A r of 1.0079 which shows that most Hydrogen atoms have a mass number of 1 but there are some with a mass number of 2. The mean (average) mass number is 1.0079

6. Relative Molecular Mass Atoms can bond together to form molecules or ionic compounds. When this happens we can add up all the Relative Atomic masses to get a Relative Molecular mass (M r ) – also called a molar mass H 2 is two hydrogen atoms combined. The Relative Molecular mass of H 2 is: 2 x 1.0079 = 2.0158 H 2 O is two hydrogen atoms and one oxygen combined. The M r (H 2 O) is: 2 x 1.0079 + 15.999 NOTE: Molecular and Atomic mass can also be called Molar mass (M). A shorthand way of writing “the molar mass of H 2 O” is M(H 2 O)

7. Calculations The mass, amount or relative atomic mass can be calculated. This is the equation: m = A r x n A r or M r can be used in this equation The A r or M r can always be worked out from a periodic table In other words we need two values before we can calculate the missing one: For example: m = 100g, A r = 1.008 gmol -1 n = 100/1.008 = 99.2 mol In other words 100g of hydrogen atoms is 99.2 moles Note – all masses MUST be measured in grams before you can use this equation This equation can also be written using the following symbol: m = Mn (M = molar mass, another way of writing A r or M r )

8. Example: Calculating an amount Joan has a 56g mass of copper sulphate. What amount does she have? What do we know?Mass (m) = 56g Molar mass (M r ) = m = Mr x n The M r can be calculated if we have a periodic table Amount (n) = Cu = 63.5 S = 32.0 4 x O = 4 x 16.0 = 64.0 Total = 159.5 gmol -1 ? 159.5 gmol -1 We can now use the equation to calculate the amount But first we need to rearrange it so that it starts with “n =” n = m/M r n = 56/159.5 n = 0.35 mol ?0.35 mol Now we add the values that we know…

9. ChemicalMolar massMassAmount O100g O2O2 5.0g Na1.2mol NaOH0.0023mol Ca(OH) 2 1.3kg KMnO 4 100mg 20g1.25mol 0.38g0.010mol

10. Using Mass, Amount and Empirical Formulae Sometimes we are given percentage mass and we are asked to turn it into a empirical formula. This involves a few steps – turn the mass into amounts, then turn the amounts into a ratio (empirical formula) 1. Find the percentage mass for each element (you may be asked to calculate this i.e. element mass/total mass) 2. Pretend that you have 100g – now your percentages convert directly to grams (mass) i.e. 12.5% = 12.5g 3. Turn the mass into moles i.e. 12.5g of Carbon = 1.04moles (n = m/Mr) 4. Find the ratio of the moles of each element (divide each amount by the smallest number, then check to see if it can be simplified any further) 5. Now put the mole ratios into a formula i.e. C:H:O = 1:2:2 therefore the empirical formula is CH 2 O 2 Note: This empirical formula can only be converted into the molecular formula if you know the total mass number of the substance

11. Some Examples - Percentages Lets say “Fred” tries to extract Ca From CaCO 3. If he has 30 grams of CaCO 3 what mass of Ca is present? 1. What percentage of CaCO 3 is Ca? - The M(CaCO 3 ) = 100gmol -1 - The M(Ca) = 40gmol -1 - Therefore 40/100 (40%) is Ca 2. Now if 40% of CaCO 3 is Ca then: - 40/100 x 30g = 12g

12. Some Examples – Percentages (Part 2) Lets say “Fred” now tries to extract Ca From CaCO 3. If he wants 18 grams of Ca what mass of CaCO 3 will he need to use? 1. What percentage of CaCO 3 is Ca? - The M(CaCO 3 ) = 100gmol -1 - The M(Ca) = 40gmol -1 - Therefore 40/100 (40%) is Ca 2. Now if 40% of CaCO 3 is Ca then: - 100/40 x 18g = 45g Note – for the mathematically challenged - Did you see the difference on this slide? – Did I want to make the number smaller or larger? For example, I want an 18g sample of Ca. If I’m extracting it from CaCO 3. The CaCO 3 is going to be a larger number than 18g. Therefore my multiplier needs to be 100/40 (the opposite to the percent)

13. Balanced Equations Balanced equations like the one below show the proportion of amounts being reacted: 2H 2 + O 2 -> 2H 2 O In other words 2 moles of H 2 reacts with 1 mole of O 2 to make 2 moles of H 2 O. These are proportions and can be changed e.g. what happen if I only have 0.2 moles of H 2 ? 0.2H 2 + 0.1O 2 -> 0.2H 2 O The proportions of each chemical are still the same (but each one is now 10x less than before)

14. Balanced Equations 2 Here is how we can use balanced equations: 2H 2 + O 2 -> 2H 2 O “Jill” has 18g of H 2 – what amount of H 2 O will be made? - First we convert the mass of the H 2 into amount (moles) n = m/M = 18/2 = 9moles - Next we work out the proportions in the balanced equation: 9H 2 + 4½O 2 -> 9H 2 O Note: If we need to, we can now calculate the other masses using: m = Mn e.g. m(H 2 O) = 2 x 9 = 18g

15. Challenges… Try these: Joe burns 100g of carbon in oxygen to make CO 2. What mass of CO 2 will he make? Julie reacts Sodium with Chlorine gas (Cl 2 ) to make 20g of NaCl. What masses of each reactant did Julie use? Dave extracts 10g of Ca from a 100g sample of limestone (impure CaCO 3 ). What percentage of the sample was pure CaCO 3 ? (What assumptions must we make?)

16. See: www.webelements.comwww.webelements.com