AQA GCSE chapter 04 calculations
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| work out the formula mass of NaCl. Atomic masses of Na=23; Cl=35.5 | Formula Mass of NaCl = 1xMass Na + 1xMass Cl = 23 + 35.5 =58.5
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| work out the formula mass of NH4NO3. Atomic masses of N=14, H=1, O=16 | 14*1 + 1*4 + 14*1 + 16*3 = 80
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| CHALLENGE work out the formula mass of (NH4)2SO4; Atomic masses of N=14; H=1; S=32; O=16 | Formula mass of (NH4)2SO4= (massN x1 + mass H x4)x2 + massS x1 + massO x4= (14x1 + 1x4)x2 + 32x1 + 16x4 = 132
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| work out the formula mass of MgSO4; Atomic masses of Mg=24; S=32; O=16 | Formula mass MgSO4 = MassMg x1 + MassS x1 + MassO x4= 24x1 + 32x1 + 16x4 = 120
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| how do you work out the formula mass of a compound? | multiply the mass of each element by the subscript that follows; then add the masses together; MgCl2
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| CHALLENGE Fe2O3 + 3 CO -> 2 Fe + 3 CO2; what mass of iron is produced when 80g of Iron(III)oxide is reacted? ( Atomic masses of Fe=56; O=16) | Formula mass Fe2O3=56x2+3x16=160; moles Fe2O3=mass/formula mass=80/160=0.5; ratio of BIG numbers says moles Fe= 2x molesFe2O3 =0.5x2=1; mass Fe=moles Fe x formula mass=1x56=56g; remember to use BIG numbers in ratio ONLY
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| 2 KClO3 -> 2 KCl + 3 O2; What mass of KCl is produced when 122.5g of KClO3 reacts completely? (Atomic masses of K=39; Cl=35.5; O=16) | Formula mass KClO3=39+35.5+3x16=122.5; form.mass KCl=39+35.5=74.5; moles KClO3=mass/formula mass=122.5/122.5=1; ratio of BIG numbers says moles KCl= moles KClO3 =1; mass KCl=moles KCl x formula mass=1x74.5=74.5g
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| CuCO3 -> CuO + CO2; What mass of Carbon Dioxide is produced when 12.4g of CuCO3 reacts completely? (Atomic masses of Cu=64; C=12; O=16) | Formula mass CuCO3=64+12+3x16=124; form mass CO2=12+16x2=44; moles CuCO3=mass/formula mass=12.4/124=0.1; ratio of BIG numbers says moles CO2=moles CuCO3 =0.1; mass CO2=moles CO2 x formula mass=0.1x44=4.4g
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| how do you work the mass of a compound, given the equation for the reaction and the mass of the reactant? | set up the calculations under the equation: first line is mass; second line is formula mass; bottom line is moles; then go down (divide mass by formula mass); across (look at the BIG numbers); and up (multiply moles by formula mass)
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| how do you work out the yield of a reaction? | yield = 100 x mass of product actually produced / maximum theoretical mass of product
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| 2 NaHCO3 --> Na2CO3 + H2O + CO2; if 16.8g of NaHCO3 are reacted and only 9.2g of Na2CO3 are produced, what is the percentage yield? | RFM(NaHCO3)=84; moles(NaHCO3)=16.8/84=0.2; moles(Na2CO3)=0.5xmoles(NaHCO3)=0.1; RFM(Na2CO3)=116; theoretical mass(Na2CO3)=0.1x116=11.6; percentage yield=100x9.2/11.6=79.3%
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| CaCO3 --> CaO + CO2; if 200t of CaCO3 are reacted and 98t of CaO are produced, what is the percentage yield? | RFM(CaCO3)=100; moles(caCO3)=200/100=2; moles(CaO)=moles(CaCO3)=2; RFM(CaO)=56; theoretical mass(CaO)=2x56=112; percentage yield=100x98/112=87.5%
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| Give 3 reasons why percentage yield are less than 100% | chemicals are lost during transfer from one place to another during separation; reaction is reversible; there may be side reactions producing other products
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| What is a limiting reactant? | The reactant that gets used up first in a reaction; the reactant that is found in the smallest number of moles
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| What is the relative atomic mass? | It is the average mass of one atom of an element compared to 1/12th of an atom of Carbon-12
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| What is the empirical formula? | he empirical formula of a compound is the simplest WHOLE number ratio of atoms of each element in the compound. (so the smallest number must be 1: CH2 is correct, C0.5H is wrong
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| How do you work out the empirical formula from the masses of each element in a compound? | set up the calculations under each element: first line is mass; second line is Ar, atomic mass; bottom line is moles (mass divided by Ar); then divide each mole by the smallest mole
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| A sample of Magnesium oxide contains 2.4g of Magnesium and 1.6g of Oxygen. What is the empirical formula? Atomic mass (Ar) of Magnesium is 24g and Ar of Oxygen is 16g | Calculate moles of Magnesium: 2.4g / 24g = 0.1 mole; Calculate moles of Oxygen: 2.4g / 24g = 0.1 mole; divide both moles by the smallest: here in each case 0.1 / 0.1 = 1; the empirical formula is Mg1O1 or simply MgO
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| A sample of Copper Chloride contains 3.15g of Copper and 3.55g of Chlorine. What is the empirical formula? (Atomic mass (Ar) of Copper is 63g and Ar of Chlorine is 35.5g | Calculate moles of Copper: 3.15g / 63g = 0.05 mole; Calculate moles of Chlorine: 3.55g /35.5g = 0.1 mole; divide both moles by the smallest: here for Cu 0.05 / 0.05 = 1 and for Cl 0.1/0.05=2; the empirical formula is CuCl2
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| What is the molecular formula? | Empirical formula is the simplest whole number ratio of atoms of each element in a compound, like CH2
Molecular formula is the ACTUAL number of atoms of each element in a compound. (so it may not necessarily be 1), like C2H4 or C3H6
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| How do you work out the molecular formula from empirical formula? | Add up the atomic masses of the atoms in the empirical formula (ie, calculate the formula mass of the empirical formula); then divide the Mr by the formula mass of the empirical formula; the number found is the is the multiple of empirical formula
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| The empirical formula of a hydrocarbon is CH2 and its Mr is 42. Work out the formula of that hydrocarbon . | Add up the atomic masses of the atoms in the empirical formula so 12x1+1x2=14g; then divide the Mr by the formula mass of the empirical formula so 42 / 14 = 3; the number found is the multiple of empirical formula so 3(CH2), the molecular formula is C3H6
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