# C3-How Much?

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• Created by: Poppy
• Created on: 14-03-13 19:23

## 3.1 The Mass Of Atoms

• Relative mass of protons and nuetrons is 1
• The atomic number of an atom it its number of protons(same as number of electrons).
• The mass number of an atom is the total number of protons and neutrons in its nucleus.
• Isotopes are atoms of the same element with different numbers of neutrons.
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## 3.2 Masses Of Atoms And Moles

• We use relative atomic masses(Ar) to compare the masses of atoms. The relative atomic mass of an element in grams is called one mole of atoms of that element.
• The relative atomic mass of an element is an average value for the isotopes of an element.
• We work out the relative formula mass of a compound by adding up the relative atomic masses of the elements in it.

E.G Calculate the Mr of calcium chloride, CaCl2

Ar of Ca=40, Ar of Cl=35.5,

So Mr =40+(35.5x2)=111

• One mole of any substance is its relative formula mass in grams. Using moles of a substance is useful when we need to work out how much of a substance reacts or how much product we will get.

E.G What is the mass of one mole of sodium hydroxide, NaOH?

Ar of Na=23, Ar of O=16, Ar of H=1, so 1 mole of NaOH=(23+16+1)g=40g

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## 3.3 Percentages and Formulae

• The relative atomic masses of the elements in a compound and its formula can be used to work out its percentage composition

E.G What is the percentage of carbon in carbon dioxide, CO2?

Ar of C=12, Ar of O=16   Mr of CO2=12+(16x2)=44

So percentage of carbon=(12/44) x 100=27.3%

• We can calculate the empirical formula given the masses or percentage composition of elements present.

E.G What is the empirical formula of the hydrocarbon that contains 80% carbon?

Carbon             Hydrogen

Mass of 100g of compound                             80                     20

Simplest ratio of atoms(mass/Ar)                80/12=6.67       20/1=20

Simplest ratio of atoms(divide by smallest)  6.67/6.67=1    20/6.67=3 so  Empirical Formula          =        CH3

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## 3.4 Equations and Calculations

Chemical equations show the reactants and products of a reaction. When balanced they show amounts of atoms, molecules or ions in the reaction. For example 2Mg+O2-->2MgO shows us two magnesium atom react with one molecule of oxygen to form 2 magnesium ions and two oxide ions...If we work in moles, the equation tells us 2 molesof magnesium atoms react with 1 mole of oxygen molecules to produce 2 moles of magnesium oxide.

This means that 48g of magnesium reacts with 32g of oxygen to give 80g of magnesium oxide (Ar of Mg=24, Ar of O=16)

Alternatively if we work in relative masses from the equation: (2 x Ar of Mg)+(2 x Ar of O) gives (2 x Mr of MgO).

Converting this to grams it becomes 2 x 24g Mg + 2 x 16g O gives 2 x40g MgO or 48g Mg+32g O gives 80g MgO (which is the sameas when we used moles)

If we have 5g of magnesium, we can work out the mass of magnesium oxideit produce using ratios:1g of Mg will produce 80/48g MgO=8.33g of MgO .If we use moles the calculation can be done like this: 1 mole of Mg produces 1 mole of MgO

5g Mg=5/24 moles of magnesium and so it will produce 5/24 moles of MgO]

The mass of 5/24 moles of MgO=5/24 x 40g= 8.33g of MgO

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## 3.5 The Yield Of A Chemical Reaction

• The yield of a chemical reaction describes how much product is made.
• The percentage yield of a reaction tells us how much product is made compared with the maximum amount that could be made.

How to calculate percentage yield

The percentage yield= amount of product collected      X100

maximum amount of product possible

The  maximum amount of product possible is calculated from the balanced equation for the reaction.

For example: A student collected  2.3g of magnesium oxide from 2.0g of magnesium.

Theoretically:2MgO+O2-->2MgO, so 48g of Mg should give 80g of MgO and so 2.0g

of Mg should give 2x 80/48=3.33g of MgO. so Percentage Yield=(2.3/3.33) x 100=69%

• It is important to maximise yield and minimise energy wasted to conserve the Earth's limited resources and reduce pollution
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## 3.6 Reversible Reactions

• In a reversible reaction the products of the reaction can react to make the original reactants.
• We can show a reversible reaction using the reversible reaction sign,
• Remember that the equation..

ammonia chlorideammonia+hydrogen chloride

AND that this is just one example of this type of reaction.

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## 3.7 Analysing Substances

• Substances added to food to improve  its qualities are calld food addtives. Food additives may be natural products or synthetic chemicals.
• Food can be checked by chemical analysis to ensure only safe,permitted additives have been used. The methods used include paper chromatography and mass spectrometer.
• Paper chromatography can be used to analyse the artaficial colours in foo. A spot of colour is put onto paper and a solvent is allowed to move through the paper. The colours ove different distances depending on their solubility.
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## 3.8 Instrumental Analysis

• Modern instrumental techniques provide fast, accurate and sensitive ways of analysing chemical substances.
• Compounds in a mixture can be separated using gas chromatography.
• Gas chromatography the mixture is carried by agas through a long column packed with particles of solid. The individual compounds travel at different speeds through the column and come out at different times. The amount of substance leaving the column in the mixture and their retention times. The retention times  can be compared with the results for known compounds in the mixture.
• Once separates, compounds can be identified using a mass spectrometer which gives further data that a computer can use quickly to identify individual compounds.
• A mass spectrometer can give the relative molecular mass of a compound. For an individual compound the peak with the largest mass correspondsto an ion with just one electron removed. This peak is called the molecular ion peak and is furthest to the on a mass spectrum.
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