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GCSE Chemistry Flash Card Questions

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C4 Chemical calculations

C4.1 Relative masses and moles

Question	Answer
Why are actual masses not used when dealing with atoms?	because they are so tiny as to be impractical
What is used instead of actual mass when dealing with atoms?	relative mass
What is the relative mass of an atom called?	the relative atomic mass
State the symbol for the relative atomic mass.	Ar
What is the standard reference point for relative atomic mass?	an atom of Carbon-12, 12C
What is the mass of an atom of Carbon-12?	12 units
Why does Carbon-12 have a relative atomic mass of 12 units?	because it has 6 protons and 6 neutrons
Where can the relative atomic mass of an atom of an element be found?	on the periodic table
Why are some of the relative atomic masses on the periodic table not whole numbers?	the numbers on the periodic table are the average atomic mass of all isotopes of the element
Why is the relative atomic mass of Chlorine 35.5 units when it is not possible to get half a proton or a neutron?	because there are two main isotopes of Chlorine that occur naturally namely Chlorine-35 and Chlorine-37
What is the relative formula mass?	the sum of the atomic masses of all the atoms in the formula
What is the symbol for relative formula mass?	Mr
Work out the Mr of sulphuric acid?	Mr(H2SO4)= (1x2) + 32 + (16x4) = 98 units
What is the relative molecular mass?	the relative formula mass of a molecule
What does the mole measure?	the amount of a substance
How is the mass of a mole worked out?	the relative atomic or formula mass expressed in grams (g)
How many atoms in a mole of atoms?	6.02 x 1023 atoms
What is the number 6.02 x 1023 called?	the Avogadro constant
State the formula that relates the number of moles to actual mass.	the number of moles = the actual mass in grams divided by relative formula mass \( n = \frac{m}{Mr} \)
How many moles in 30g of sulphuric acid?	Mr(H2SO4)= (1x2) + 32 + (16x4) = 98 units \( n = \frac{30}{98} \) = 0.31 moles

C4 Chemical calculations

C4.2 Equations and calculations

Question	Answer
Write the balanced equation for making hydrogen chloride molecules.	H2 + Cl2 → 2HCl
What are the reactants in the equation H2 + Cl2 → 2HCl?	H2 and Cl2 are the reactants
What are the products in the equation H2 + Cl2 → 2HCl?	HCl is the only product
In terms of molecules, what does the equation H2 + Cl2 → 2HCl say?	1 molecule of H2 and 1 molecule of Cl2 makes 2 molecules of HCl
In terms of moles, what does the equation H2 + Cl2 → 2HCl say?	1 mole of H2 and 1 mole of Cl2 makes 2 moles of HCl
In terms of moles, C2H6 + \( 3 \frac{1}{2} \) O2 → CO2 + 3H2O say?	1 mole of C2H6 and \( 3 \frac{1}{2} \) moles of O2 makes 1 mole of CO2 and 3 moles of H2O
What is the mass of 1 mole of H2?	mass of 1 mole of H2 = 2 x 1 = 2g
What is the mass of 1 mole of Cl2?	mass of 1 mole of Cl2 = 2 x 35.5 = 71g
What is the mass of 1 mole of HCl?	mass of 1 mole of HCl = (1 + 35.5) = 36.5g
What mass of HCl will 1 mole of H2 and 1 mole Cl2 make?	this will make 2 moles of HCl so 2 x 36.5 = 73g
What mass of chlorine gas is needed to convert 100.0g of sodium hydroxide to bleach? 2NaOH + Cl2 → NaOCl + NaCl +H2O	• mass of 1 mole of NaOH = 23 + 16 + 1 = 40g • mass of 1 mole of Cl2 = 35.5 x 2 = 71g • number of moles of NaOH in 100g = \( \frac{100}{40} \) = 2.5 moles • the equation shows that 2 moles of NaOH reacts with 1 mole of Cl2 • so 2.5 moles of NaOH require \( \frac{2.5}{2} \) = 1.25 moles of Cl2 • mass of 1.25 moles of Cl2 = 1.25 x 71 = 88.75g (4sf) • the question gives values to 4sf so it is appropriate to give the answer to 4sf

C4 Chemical calculations

C4.3 From masses to balanced equations

What is the stoichiometry of a reaction?	the ratio of the number of moles of each reactant and product
Write the formula for sodium nitrate?	NaNO3
What happens to NaNO3 when it is heated?	it decomposes to form NaNO2 and O2
Write the equation for the thermal decomposition (decomposition by heat) ofNaNO3?	sodium nitrate → sodium nitrite + oxygen
What is NaNO2?	sodium nitrite
What is the mass of oxygen is formed when 8.5g of NaNO3 is heated fully leaving 6.9g of NaNO2?	mass of oxygen = 8.5 - 6.9 = 1.6g
How many moles are in 8.5g of NaNO3?	Mr of NaNO3 = [23 + 14 + (3x16)]= 85 number of moles of NaNO3 in 8.5g = \( \frac{8.5}{85} \) = 0.1 moles
How many moles are in 6.9g of NaNO2?	Mr of NaNO2 = [23 + 14 + (2x16)]= 69 number of moles of NaNO2 in 6.9g = \( \frac{6.9}{69} \) = 0.1 moles
How many moles are in 1.6g of O2?	Mr of O2 = [2x16]= 32 number of moles of O2 in 1.6g = \( \frac{1.6}{32} \) = 0.05 moles
What is the ratio of NaNO3 : NaNO2 : O2	0.1 : 0.1 : 0.05 multiply by 20 to get whole numbers 2 : 2 : 1
Write the balanced equation for the thermal decomposition of sodium nitrate.	the ratios calculated above show the balanced quantities in the equation so the balanced equation is 2NaNO3 → 2NaNO2 + O2
What does it mean when a reactant is in excess?	there will be some of thia reactant left over when the rest of the reactants are used up
What is the limiting reactant?	the reactant that first gets used up is the limiting reactant
What would be used up completely when reacting magnesium ribbon with hydrochloric acid if hydrochloric acid is the limiting reactant?	the hydrochloric acid
What would be left over when reacting magnesium ribbon with hydrochloric acid if hydrochloric acid is the limiting reactant?	some magnesium ribbon
Write the balanced equation for magnesium reacting with hydrochloric acid.	Mg + 2HCl → MgCl2 + H2
If 4.8g of Mg reacts with 7.3g of HCl, which is the limiting reactant?	number of moles of Mg in 4.8g = \( \frac{4.8}{24} \) = 0.2 moles number of moles of HCl in 7.3g = \( \frac{7.3}{36.5} \) = 0.2 moles the balanced equation shows that 1 mole of Mg reacts with 2 moles of HCl this means that 0.2 moles of Mg need 0.4 moles of HCl to react fully with only 0.2 moles of HCL, the HCL is the limiting reactant and Mg is in excess

C4 Chemical calculations

C4.4 The yield of a chemical reaction (single science only)

Question	Answer
In the reaction A +2B → C, will 1 mole of A produce 1 mole of C in real life?	No, it will not
In the above reaction, why will 1 mole of A not produce 1 mole of C in real life?	three possible reasons are that some of A may be lost in handling, some of A may stick to the side of the container, and the substance A may contain impurities
What is meant by percentage yield?	the percentage yield of a chemical reaction tells you how much actual product is made compared with the maximum theoretical amount that could be made
State the formula for yield	percentage yield = \( \frac{actual\; mass\; of\; product\; produced}{maximum\; theoretical\; mass\; of\; product\; produced} \)
If the maximum mass of product predicted for a reaction is 1200kg, but only 900kg are produced in reality, what is the percentage yield?	percentage yield = \( \frac{900}{1200} \times 100 \) = 75%
When Calcium Carbonate is heated to produce Calcium Oxide, why is the yield not 100%?	some limestone is lost as dust during crushing, some limestone is lost as dust in the rotating kiln, and the limestone is not 100% Calcium Carbonate as their are other compounds present
List the 5 main reasons chemical reactions do not have 100% yield	• the reaction may be reversible • some reactants may produce unexpected or unwanted products follow different paths • some reactant may be lost in handling or left in the apparatus • the reactants may not be pure • not all the desired product may be removed from the reaction mixture
If that 300 tonnes of CaCO3 produces 142 tonnes of CaCO, what is the percentage yield?	CaCO3(s) → CaO(s) + CO2(g) Mr of CaCO3 = [40 + 12 + (16 x 3)]= 100 Mr of CaO = [40 + 16)]= 56 Thus 100 tonnes would produce 56 tonnes in a perfect world And 300 tonnes would produce 168 tonnes in a perfect world percentage yield = \( \frac{actual\; mass\; of\; product\; produced}{maximum\; theoretical\; mass\; of\; product\; produced} \) percentage yield = \( \frac{142}{168} \) x 100 = 84.5%

C4 Chemical calculations

C4.5 Atom economy (single science only)

Question	Answer
Who are the main people in the designers of chemical factories and plants?	chemical engineers
Name some criteria for factory and plant design	•   safety •   reduce energy use •   reduce waste of raw materials
What is the other term for atom economy?	atom utilisation
What is atom economy?	the atom economy is a measure of the extent to which the atoms in the starting materials (reactants) end up in the desired product
What happens to the atoms not in the desired products?	they end up in other products sometimes called waste products
What happens to waste products?	first, the company will try to find uses for waste products, and if this is not possible, then they are disposed of safely
State the equation for percentage atom economy	percentage atom economy = \( \frac{relative\; formula\; mass\; of\; the\; desired\; product\; from\; equation} {sum\; of\; the\; relative\; formula\; masses\; of\; the\; reactants\; from\; equation} \) x 100%
What reactions produce the ideal 100% "atom economy"?	all the atoms in the reactants appear in the products
State an equation with the ideal 100% "atom economy"	N2(g) + 3H2(g) → 2NH3(g)
The reaction below is used to produce PbO      2PbS(s) + 3O2(g) → 2PbO(s) + 2SO2(g) Calculate the percentage economy for PbO	percentage atom economy = \( \frac{relative\; formula\; mass\; of\; the\; desired\; product\; from\; equation} {sum\; of\; the\; relative\; formula\; masses\; of\; the\; reactants\; from\; equation} \) x 100% = \( \frac{M_r(2PbO)}{[M_r(2PbS) + M_r(3O_2)]} \) x 100% = \( \frac{2 \times (207 + 16)}{[2 \times (207 + 32)] + [3 \times (16 \times 2)]} \) x 100% = \( \frac{446}{(478 + 96)} \) x 100% = 77.7%
Is SO2 a problem as a product in the above equation?	yes, it is a serious pollutant causing acid rain
What is done with the SO2?	it is reacted with a base such as calcium oxide or calcium carbonate to produce calcium sulphate
What is done with the calcium sulphate produced?	it is used in the construction industry or dumped in landfill

C4 Chemical calculations

C4.6 Expressing concentrations

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C4 Chemical calculations

C4.7 Titrations (single science only)

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C4 Chemical calculations

C4.8 Titration calculations (single science only)

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C4 Chemical calculations

C4.9 Volumes of gases (single science only)

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