# C2.3 How much?

?
• Created by: Twins 1&2
• Created on: 09-02-14 16:06

## 3.1 The mass of atoms

MASS NUMBER

• the total number of protons and neutrons in a nucleus.

ATOMIC NUMBER

• the number of protons in the nucleus of an atom, which is characteristic of a chemical element and determines its place in the periodic table.

ISOTOPES

• each of two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei
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## 3.2 Masses of atoms and moles

Relative atomic mass of an element in grams is called one mole of atoms of the element.

To find the relative formula mass (or Mr) of a substance, you add together therelative atomic mass for all the atoms shown in its chemical formula.

Remember that the values for relative atomic masses (Ar) are given in the periodic table. So, although it helps to learn some of the common values, you do not have to do this.

## Example 1

What is the relative formula mass of water, H2O?

(Ar of H = 1, Ar of O = 16)

Mr of H2O = 1 + 1 + 16 = 18

## The mole

The unit for amount of substance is called the mole, shown as mol. One mole of atoms, ions or molecules is around 6 × 1023. This is the same number as the number of carbon atoms in 12 g of carbon.

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## 3.3 Percentages and formulae

• Percentage of any element = Relative atomic mass (A r) / relative formula mass (M r) x 100
• Worked example;
• % of carbon in Carbon dioxide (CO2)?
• Ar of C = 12  Ar of O = 16
• Mr of CO2 = 12 + (16x2) = 44
• % of C = (12/44) x 100 = 27.3%

Working out formula of compounds % composition

Empirical formula (simplest ratio of atoms or ions in compound)

• Empirical formula(to give ratio) = mass / Ar
• Worked example;
• Empirical formula of Hydrocarbon with 80% carbon?
• Mass in 100g = 80g (carbon) 20g (hydrogen)
• Ratio/mole of atoms = 80/12= 6.67    20/1=20
• simplest ratio = 6.67/6.67=1   20/6.67=3
• EMPIRICAL FORMULA = CH3
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## 3.4 Equations and calculations

Worked example

Calculate the mass of calcium oxide that can be made from 10g of calcium carbonate n this reaction;

CaCO3 -- CaO + CO2 (Ar of Ca = 40, Ar of O = 16, Ar of C = 12)

CaCO3 =40 +12+(16x3) = 100

CaO = 40 + 16 = 56

CO2 = 12 + (16 x 2 ) =44

10g of CaCO3 = 100/10 = 10g

CaO produced = 56/10 = 5.6g

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## 3.5 Yield of a chemical reaction

• % Yield =    amount collected            x 100%
•        maximum amount possible
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## 3.6 Reversible reactions

If the products of a recation can react to produce the reactants then that reaction is called a reversible reaction.

An example would be; 6 of 8

## 3.7 Analyzing substances

Substances are added to food to mporve its qualities are called food additives. these may be natural or synthetic.

Foods can be checked by chemical analysis to ensure only safe permitted addicitves have been used. Methods include;

• chromatography paper
• mass spectrometry

Paper chromatorgraphy can be used to analyse to the artificial colours in fodd. A spot of colour is put onto paper and a solvent is allowed to move through the paper. The colours move different distances depeding on ther solubility.

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## 3.8 Instrumental analysis

Samples to be analysed are often mixtures of different compounds and they need to be seperated before they can be identified by instrumental techniques. To do this chemists have a developed a technique called gas chromatography-mass spectrometry (GC-MS). They use gas chromatography to seperate the compounds that are easily vapourised.

Seperation technique:

• 1. Sample mixture is vapourised
• 2. A carrier gas moves the vapour through the column (usually helium)
• 3. The compounds have different attractions to the material inside the column. The compounds with stronger attractions to the column will take longer to get through the column. We say they have a longer retention time. The compounds that take a shorter amount of time to get throught the column have a shorter retention time.

To identify compounds we compare the chromotograph (the record of seperated compounds taken when the compounds leave the column) with known substances. Must take place in same conditions to compare retention times.

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