mass = moles x m r moles = concentration x volume to convert volume from cm 3 → dm 3, divide by...
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![Page 1: Mass = Moles x M r Moles = Concentration x Volume To convert volume from cm 3 → dm 3, divide by 1000](https://reader033.vdocuments.site/reader033/viewer/2022061313/5697bfde1a28abf838cb2488/html5/thumbnails/1.jpg)
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Mass = Moles x Mr
Moles = Concentration x Volume
To convert volume fromcm3 → dm3, divide by 1000
![Page 3: Mass = Moles x M r Moles = Concentration x Volume To convert volume from cm 3 → dm 3, divide by 1000](https://reader033.vdocuments.site/reader033/viewer/2022061313/5697bfde1a28abf838cb2488/html5/thumbnails/3.jpg)
Example 1: 3.00 g of lawn sand containing an iron(II) salt was shaken with dilute sulphuric acid
The resulting solution needed 25.00 cm3 of 0.0200 mol dm-3 KMnO4 to oxidise the Fe2+ ions in the solution to Fe3+ ions
Use this information to calculate the percentage by mass of Fe2+ ions in this sample of lawn sand
![Page 4: Mass = Moles x M r Moles = Concentration x Volume To convert volume from cm 3 → dm 3, divide by 1000](https://reader033.vdocuments.site/reader033/viewer/2022061313/5697bfde1a28abf838cb2488/html5/thumbnails/4.jpg)
1) Equation:
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
2) Moles of MnO4- = Conc x Vol
= 0.200 x (25 ÷ 1000) = 5 x 10-3 mol
3) Moles of Fe2+ = Moles of MnO4- x 5 (Ratio 1:5)
= 5 x 10-3 x 5 = 0.025 mol
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4) Mass of Fe2+ = Moles x Mr
= 0.025 x 55.8 = 1.40 g
5) % by mass of Fe2+ = (1.40 ÷ 3.00) x 100 = 46.7 %
The most important part is to write the equation for reaction as it allows you to work out the ratio (step 3)
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Example 2: In a titration, 0.321 g of a moss killer reacted with 23.60 cm3 of acidified 0.0218 mol dm-3 K2Cr2O7 solution
Calculate the percentage by mass of iron in the moss killer
Assume that all of the iron in the moss killer is in the form of iron(II)
![Page 7: Mass = Moles x M r Moles = Concentration x Volume To convert volume from cm 3 → dm 3, divide by 1000](https://reader033.vdocuments.site/reader033/viewer/2022061313/5697bfde1a28abf838cb2488/html5/thumbnails/7.jpg)
1) Equation:
Cr2O72- + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+
2) Moles of Cr2O72- = Conc x Vol
= 0.0218 x (23.60 ÷ 1000) = 5.145 x 10-3 mol
3) Moles of Fe2+ = Moles of Cr2O72- x 6 (Ratio 1:6)
= 5.145 x 10-3 x 6 = 3.087 x 10-3 mol
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4) Mass of Fe2+ = Moles x Mr
= 3.087 x 10-3 x 55.8 = 0.172 g
5) % by mass of Fe2+ = (0.172 ÷ 0.321) x 100 = 53.6 %
The most important part is to write the equation for reaction as it allows you to work out the ratio (step 3)
![Page 9: Mass = Moles x M r Moles = Concentration x Volume To convert volume from cm 3 → dm 3, divide by 1000](https://reader033.vdocuments.site/reader033/viewer/2022061313/5697bfde1a28abf838cb2488/html5/thumbnails/9.jpg)
Example 3: A 1.27g sample of impure iron was reacted with an excess of dilute sulphuric acid
The solution formed was made up to 250 cm3
A 25.0 cm3 sample of this solution reacted completely with exactly 19.6 cm3 of a 0.0220 mol dm-3 solution of potassium manganate(VII)
Calculate the percentage by mass of iron in the sample
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1) Equation:
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
2) Moles of MnO4- = Conc x Vol
= 0.0220 x (19.6 ÷ 1000) = 4.31 x 10-4 mol
3) Moles of Fe2+ = Moles of MnO4- x 5 (Ratio 1:5)
= 4.31 x 10-4 x 5 = 2.16 x 10-3 mol in 25 cm3
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4) Moles of Fe2+ in 250 cm3 = 2.16 x 10-3 x 10 = 2.16 x 10-2 mol
5) Mass of Fe2+ = Moles x Mr
= 2.16 x 10-2 x 55.8 = 1.21 g
6) % by mass of Fe2+ = (1.21 ÷ 1.27) x 100 = 95.3 %
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Example 4: A solution of iron(II) sulfate was prepared by dissolving 10.00 g of FeSO4.7H2O (Mr = 277.9) in water and making up to 250 cm3 of solution
The solution was left to stand, exposed to air, and some of the iron(II) ions became oxidised to iron(III) ions
A 25.0 cm3 sample of the partially oxidised solution required 23.70 cm3 of 0.0100 mol dm-3 potassium dichromate(VI) solution for complete reaction in the presence of an excess of dilute sulfuric acid
Calculate the percentage of iron(II) ions that had been oxidised by the air
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1) Equation:
Cr2O72- + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+
2) Moles of Cr2O72- = Conc x Vol
= 0.0100 x (23.70 ÷ 1000) = 2.37 x 10-4 mol
3) Moles of Fe2+ = Moles of Cr2O72- x 6 (Ratio 1:6)
= 2.37 x 10-4 x 6 = 1.42 x 10-3 mol in 25 cm3
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4) Moles of Fe2+ in 250 cm3 = 1.42 x 10-3 x 10 = 1.42 x 10-2 mol
5) Original moles of Fe2+ = Mass ÷ Mr
= 10.00 ÷ 277.9 = 0.036 mol
6) Moles of Fe2+ oxidised = Original - Fe2+ remaining = 0.036 - 1.42 x 10-2
= 0.0218 mol
7) % by mass of Fe2+ oxidised = (0.0218 ÷ 0.036) x 100 = 60.6 %