Calculations in Chemistry- part 2

Molar volume What is the mass of: 600cm^3 of Ammonia gas NH 3 at RTP? 0.43g 1000mL of Methane CH 4 gas at RTP? 0.67g 4800 cm^3 Oxygen gas O 2 at RTP? 6.40g cm^3 of Nitrogen dioxide NO 2 gas at RTP? 19.17g

Molar volume What is the volume of: moles of hydrogen H 2 at RTP? 72cm^3 8g of oxygen O 2 gas at RTP? 6000 cm^3 120g of sulphur dioxide SO 2 gas at RTP? cm^3 5.6 g of Nitrogen N 2 gas at RTP? 4800 cm^3

Reacting volumes Gases react according to the mole ratio in a balanced equation. Volume of a gas is proportional to the moles. 2H 2(g) + O 2(g)  2H 2 O (g) In the above reaction, hydrogen, oxygen and water vapour are in the ratio 2:1:2. If 100 cm^3 of hydrogen were completely reacted, 50cm^3 oxygen are used and 100cm^3 water vapour are produced.

Nitrogen and Hydrogen react according to the equation. N 2(g) + 3H 2(g)  2NH 3(g) If 150cm^3 of ammonia are formed, what will be the volume of nitrogen and hydrogen reacted? N 2 :NH 3 = 1:2 So nitrogen reacted is 75cm^3 N 2 : H 2 = 1:3 So hydrogen reacted is 225cm^3

In the following reaction 200 cm^3 of Sulphur dioxide gas was mixed with 175cm^3 oxygen. 2SO 2(g) + O 2(g)  2SO 3(g) (a) Balance the equation (b) Which gas will be completely used up? SO 2 (limiting agent) (c) What is the volume of SO 3 produced? 200cm^3 (d) Name gases present when the reaction finished? SO 3 and remaining O 2 (e) What is the final volume of the gaseous mixture? 200cm^3 (SO 3 )+ 75cm^3 (O 2 ) = 275cm^3

Methane burns in air according to the following equation: CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (g) In an experiment, 600cm^3 of methane were burnt in 1500cm^3 of O 2. (a) Which chemical is the limiting reagent? Methane is the limiting agent What is the volume of CO2 and H2O are produced? CO 2 = 600cm^3 H 2 O = 2 X 600 = 1200cm^3 What volume of unreacted gas is left? Unreacted gas is Oxygen = 1500 – 1200 = 300cm^3 What is the final volume at the end of the experiment? Final volume contains CO 2, H 2 O and excess O = 2100cm^3

% Yield IGCSE Chemistry

A student was asked the following calculation: Copper metal is made from copper(II) oxide by heating it with carbon powder. The equation is as follows: 2CuO + C  2Cu + CO2 8.0g of copper (II) oxide was heated with 4.0g of carbon powder. (a) How many moles of copper(II) oxide was used? (b) How many moles of carbon were mixed? (c) What is the limiting reagent? (d) What mass of copper is made in this calculation? (e) How many g of the other reagent is also left at the end of the reaction? (f) How many moles of CO2 were produced? (g) What is the volume of this CO2 at room temperature and pressure?

The student made calculations and made the following results. (a) moles of copper(II) oxide = 0.1 (b) moles of carbon mixed = 0.33 (c) Limiting agent is Copper(II) oxide (d) mass of copper made = 6.4g (e) g of other reagent(excess) left = 3.36g (f) Moles of CO2 produced = 0.05 (g) Volume of CO2 produced = 0.05 x = 1200 cm3

The teacher asked the student about the expected amount of copper. The student read the previous calculations and said 6.4g The teacher then asked the student to conduct an experiment using 8.0g of copper(II) oxide and 4.0g of carbon. The student conducted the experiment using the following set up What is left in the test tube when the experiment is over? What the student might have seen with lime water? Why? What can be the reason the test tube is slanted slightly down?

The student separated copper produced from remaining carbon and weighed. She found that the weight is smaller than what she calculated. That is 4.8g Why you think the amount of copper is smaller than what she calculated? What you call the amount of product calculated and the amount of product really produced in an experiment? If the formula to calculate % yield is: find out the percentage yield of copper.

Try this question now: Zinc displaces copper from copper(II) sulphate by displacement reaction according to the following equation: CuSO4 + Zn  ZnSO4 + Cu In a reaction 3.2g of copper(II) sulphate was mixed with 3.0g of Zinc. (a) How many moles of copper(II) sulphate are present in 3.2g? 0.02 moles (b) How many moles of Zinc is present in 3.0g? moles What chemical is the limiting agent? Copper(II) sulphate is the limiting agent

How many moles of the other chemical is left at the end of the reaction? – = moles of Zinc What mass is it? x 65g = 1.69g What mass of copper is produced? (theoretical yield) 0.02 x 64g = 1.28 g In an experiment, 1.10g of copper was produced (Experimental/actual yield). Find out the % yield of copper

FORMULA IGCSE Chemistry

Formula Two important types of formulae in Chemistry are: 1. Empirical formula This shows which elements are present in a compound and the ratio of various atoms in one molecule. E.g Empirical formula of Benzene is CH 2. Molecular formula Molecular formula shows actual number of various atoms in a molecule. E.g Molecular formula of Benzene is C 6 H 6

When you multiply Empirical formula with a number, you get the molecular formula. In the example of benzene, when you multiply Empirical formula with 6, you get the molecular formula (CH) X 6  C 6 H 6 You can calculate the number if you know the empirical formula and molecular mass. First find the empirical formula mass by adding various atomic masses. Then divide the molecular mass with empirical formula mass, you get the number.

Molecular formula Molecular mass will be given in the question. In the example of Benzene, molecular mass is given as 78 and empirical formula is CH Find Empirical formula mass first = = 13 Divide molecular mass with empirical formula mass to get number (n) 78/13 = 6 Now multiply Empirical formula with 6 you get molecular formula (CH) X 6 = C 6 H 6

How to find the empirical formula? To find the empirical formula, you need the mass (g) or (%) of different element. Step 1:convert mass (%) in to moles by dividing with molar mass of the element. Step 2:Find the mole ratio by dividing each moles calculated in step one with the smallest mole. Step 3: Round off the moles to the nearest whole number and write the empirical formula.

Example A hydrocarbon contains 92.3% C and 7.7% H. Its molecular mass is 78. find the empirical formula and molecular formula of the compound. Empirical formula Empirical formula = CH

Now find out n 78/13 = 6 Molecular formula is (CH) X 6 C 6 H 6

Another example Determine the empirical formula of the compound containing 37.5% C, 12.5% H, and 50.0% O by weight. Deduce the molecular formula of the compound if the molecular mass is 64.

Empirical formula mass = = 32 n = molecular mass/empirical formula mass 64/32 = 2 Molecular formula is (CH 4 O) X 2 C 2 H 8 O 2

END OF PART 2