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GCSE Calculations

AQA GCSE chapter 04 calculations

work out the formula mass of NaCl. Na=23; Cl=35.5 Formula Mass of NaCl = 1xMass Na + 1xMass Cl = 23 + 35.5 =58.5
work out the formula mass of NH4NO3 14*1 + 1*4 + 14*1 + 16*3 = 80
CHALLENGE work out the formula mass of (NH4)2SO4; 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
work out the formula mass of MgSO4; Mg=24; S=32; O=16 Formula mass MgSO4 = MassMg x1 + MassS x1 + MassO x4= 24x1 + 32x1 + 16x4 = 120
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
CHALLENGE Fe2O3 + 3 CO -> 2 Fe + 3 CO2; what mass of iron is produced when 80g of Iron(III)oxide is reacted? (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
2 KClO3 -> 2 KCl + 3 O2; What mass of KCl is produced when 122.5g of KClO3 reacts completely? (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
CuCO3 -> CuO + CO2; What mass of Carbon Dioxide is produced when 12.4g of CuCO3 reacts completely? (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
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)
What is the mass concentration of 3.56x10^-3 moles of H2SO4 dissolved in 25cm3 of solution? RMM H2SO4=2x1.0+32.1+4x16.0=98.1 ;Mass= 3.56x10^-3 x 98.1 = 0.349g ; Concentration= 0.349 / (25/1000) = 13.96 g.dm-3 ; Have you remembered the units at each stage?
how do you work out the yield of a reaction? yield = 100 x mass of product actually produced / maximum theoretical mass of product
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%
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%
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
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
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
Created by: ursulinechem2