Formulae of Carbon Compounds

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Clarissa Newman
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1 Formulae of Carbon Compounds Empirical Formula The empirical formula just shows the simplest ratio of the atoms present. For example ethane: 3g of ethane contains 2.4g of carbon, Ar = 12.0 and 0.6g of hydrogen Ar = 1.0. What is the empirical formula. Number of moles of carbon = 2.4 / 12.0 = 0.2 mol of carbon Number of moles of hydrogen = 0.6 / 1.0 = 0.6 mol of hydrogen 0.2 / 0.2 = / 0.2 = 3 The ratio is 1:3 Therefore the empirical formula for ethane is CH 3

2 Formulae of Carbon Compounds The Molecular Formula The molecular formula shows the numbers of each atom in the molecule. It is found from: The empirical formula The relative molecular mass of the empirical formula The relative molecular mass of the molecule For example: The empirical formula of ethane is CH3 and this group has a molecular mass of The molecular mass of ethane is 30.0, which is 2x15.0. Therefore there must be two units to the molecular formula in ever molecule of ethane. The molecular formula is therefore: (CH3) 2 or C 2 H 6

3 Concentrations of Solutions In everyday life the concentrations of solutions may have several different units. Ingredients labels show units such as g per 100ml, g/l, mg/ml, and % vol. Different units can be confusing and could lead to misunderstanding. In chemistry, the unit of concentration is mole per cubic decimetre, written as mol dm 3. This often shortened to M, which is pronounced as molar. But note that strictly speaking, the word molar really means divided by the number of moles. The concentration of a solution measured in mol dm 3 is its molarity. Calculating concentration Concentration is calculated using this equation: where c = concentration in mol dm 3 n = amount of substance in mol V = volume in dm 3

4 A standard solution is a solution with an accurately known concentration. It is usual to use a volumetric flask when you make a standard solution. The most commonly used size is 250cm 3. It is important to divide this volume by 1000 to convert it into dm 3 before using it in a calculation.

5 Using concentrations Find a required volume You may need to calculate the volume of solution required in chemical reactions. The equation is rearranged like this: Remember that the volume calculated will be in dm3, not cm3. You will need to Remember that the volume calculated will be in dm3, not cm3. You will need to multiply your answer by 1000 to get cm3.

6 Finding a required mass You need to calculate the mass of solid required when you make up your own solutions. The calculation is in two parts. 1 Calculate the amount of substance needed from the intended volume and concentration 2 Calculate the mass of substance needed from its amount and M r

7 Percentage Yield The yield of a chemical reaction is the amount of product formed. You can work out the theoretical yield using a reacting mass calculation. This is the maximum possible amount of product from the reactants used. In practice you are unlikely to get the theoretical yield. The actual yield is usually much less than the theoretical yield. There are several reasons for this, including: the reactants may be impure the reaction may not go to completion some of the product may be left in the container it may be difficult to purify the product. In a laboratory situation this might just be annoying, as you will have to use more of each reactant than you calculated. But in an industrial situation more raw materials and energy will be used, more waste will be produced, and it could cost a lot of money.

8 Calculating percentage yield The percentage yield tells you how close to the theoretical yield you have got. The higher the percentage yield, the closer you are to the theoretical yield. Here is the equation for calculating percentage yield: The percentage yield would be 100% if all the reactants were converted into products, and there were no losses during processes such as pouring and filtering. The theoretical yield for a certain method to obtain copper(ii) sulphate crystals is 2.0g. The actual yield obtained was 1.8g. What was the percentage yield?

9 The Haber Process Ammonia is manufactured from nitrogen and hydrogen using the Haber Process. Over one hundred million tonnes of ammonia is manufactured worldwide each year. About 85% of this is used to make fertilizers. Without the Haber Process it would be very difficult to produce enough food for everyone. Yet under typical conditions the percentage yield of ammonia is only about 15%. The reaction is reversible and does not go to completion:

11 Limiting reactant The reactants in a chemical reaction are often mixed in different from the ones in the balanced equation. Where there are two reactants, this means that one of them will be in excess. It will not all be used up in the reaction. The other reactant is the limiting reactant. It will be completely used in the reaction and so determines the theoretical yield. For example, iron reacts with Sulphur to produce iron(ii) Sulphide: One mole of iron reacts with one mole of Sulphur to produce one mole of iron(ii) Sulphide. If more than one mole of iron is mixed with one mole of Sulphur, the iron will be in excess. The Sulphur will be the limiting reactant. No matter how much excess iron is added, no more than one mole of iron(ii) Sulphide can be produced from one mole of Sulphur.

NOTES Mole Concept Chapter 3

Chapter 3 Vocabulary: NOTES Mole Concept Chapter 3 average atomic mass- Avogadro's Numberchemical equationempirical formula- Haber process- the weighted average mass of the atoms in a naturally occurring