Chapter 10: Mass and Moles

10.1-Relative Formula Mass

Compounds have a relative formula mass, M

• If you have a compound like MgCl2 then it has a relative formula mass, M, which is just the relative atomic masses of all the atoms in the molecular formula added together
• Calculating relative formula mass is straight forward enough, but things can get a bit more confusing when you start working the percentage compositions of compounds

You can calculate the % mass of an element in a compounds

• Percentage mass of an element in a compounds = Ar x number of atoms of that element / Mr of the compounds x 100
• These questions may involve recurring decimals but you will always have a calculator so you should be able to work it out

10.2-The Mole

The mole is simply the name given to an amount of a substance

• One mole of any substance is just an amount of that substance that contains an Avogadro number of particles-so 6.02 x 10,23, particles
  • The particles could be atoms, molecules, ions or electrons
• The mass of that number of atoms or molecules of any substance is exactly the same number of grams as the relative atomic mass or relative formula mass of the element or compound
  • In other words, one mole of atoms or molecules of any substance will have a mass in grams equal to the relative formula mass for that substances

Nice formula to find the number of moles in a given mass

• Number of moles = mass in g(of an element or compound) / M1(of the element of compound)
• You can rearrange the equation above using this handy formula triangle
• You could use it to find the mass of a known number of moles of a substance, or to find the M1 of a substance from a known mass and number of moles
• Just cover up the thing you want to find with your finger and write down what’s left showing

10.3-Conservation of Mass

In a chemical reaction, mass is always conserved

• During a chemical reaction no atoms are destroyed and no atoms are created
• This means there are the same number and types of atoms on each side of a reaction equation
• Because of this, no mass is lost or gained, we say that mass is conserved during a reaction
• By adding up the relative formula masses of the substances on each side of a balanced symbol equation, you can see that mass is conserved
  • The total M1 of all the reactants equals the total M1 of the products

If the mass seems to change, there’s usually a gas involved

• In some experiments you might observe a change of mass of an unsealed reaction vessel during a reaction, there’s two explanations for this
  • If the mass increases it’s probably because one of the reactants is a gas that’s found in air and all the products are solids, liquids or aqueous
  • Before the reaction, the gas is floating around in the air, It’s there but it’s not contained in the reaction vessel so you can’t account for its mass
  • When the gas reacts to form part of the product, it becomes contained inside the reaction vessel so the total mass of the stuff inside the reaction vessel increases
  • If the mass decreases, it’s probably because one of the products is a gas and all the reactants are solids, liquids or aqueous
  • Before the reaction, all the reactants are contained in the reaction vessel

10.4-The Mole and Equations

You can use moles to calculate masses in reactions

• Remember those balanced equations, well the big numbers in front of the chemical formulas of the reactants and products tell you how many moles of each substance takes part or is formed during the reaction
• You can use this to balance equation and predict atomic masses

You can balance equations using reacting masses

• If you know the masses of the reactants and products that took part in a reaction, you can work out the balanced symbol equation for the reaction
  • Divide them mass of each substance by its relative formula mass to find the number of moles
  • Divide the number of moles of each substance by the smallest number of moles in the reaction
  • If any of the numbers aren’t whole numbers, multiply all the numbers by the same amount so that they all become whole numbers
  • Write the balanced symbol equation for the reaction by putting these numbers in front of the chemical formulas

Practice Questions

• How many moles of Ca3(PO4)2 are in 57.82 grams?

• Ca3(PO4)2 = 310.18g/mol

  • 0.1864 moles

• How many grams of Zn(OH)2 are in 3.67 moles?

• molar mass:Zn(OH)2 = 99.41g/mol

  • 365 moles

• When 12 g of carbon is burned in air, 44 g of carbon dioxide is produced. What mass of carbon is needed to produce 11 g of carbon dioxide?

• 12 g of carbon makes 44 g of carbon dioxide.

• 12 ÷ 44 g of carbon will make 1 g of carbon dioxide.

  • Therefore, to make 11 g of carbon dioxide, you will need 11 × (12 ÷ 44 g) = 3 g of carbon.)

• When 5.0 g of calcium carbonate is decomposed by heating, it produces 2.2 g of carbon dioxide. What mass of calcium carbonate is needed to produce 8.8 g of carbon dioxide?

• 5 g of calcium carbonate makes 2.2 g of carbon dioxide.

• 5 ÷ 2.2 g of calcium carbonate will make 1 g of carbon dioxide.

  • Therefore, to make 8.8 g of carbon dioxide, you will need 8.8 × (5 ÷ 2.2 g) = 20 g of carbon.

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