# GCSE Chemistry

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1. the Law of Conservation of Mass" definition states
that mass cannot be created or destroyed, but changed into different forms.So, in a chemical change, the total mass of reactants must equal the total mass of products.By using this law, together with atomic and formula masses, you can calculate the quantities of reactants and products involved in a reaction and the simplest formula of a compound
2. The EMPIRICAL FORMULA of a compound can be worked out by knowing the exact masses of the elements that combine to form a given mass of a compound.The empirical formula of a compound is the
simplest whole number ratio of atoms present in a compound.  Here the word 'empirical' means from experimental data
3. It is found that 207g of lead combined with 32g of sulphur to form 239g of lead sulphide. From the data work out the formula of lead sulphide. (Relative atomic masses: Pb = 207 and S = 32)
In this case it easy to see that by the atomic mass ratio, 239 splits on a 1 to 1 basis of 1 atom of lead to 1 atom of sulphur (1 x 207 to 1 x 32 by mass), so the formula is simply PbS
4. t is found that 207g of lead combined with oxygen to form 239g of a lead oxide. From the data work out the formula of the lead oxide. (Relative atomic masses: Pb = 207 and O = 16)
In this case, you first have to work out the amount of oxygen combined with the lead. This is 239 - 207 = 32. In atomic ratio terms, the 207 is equivalent to1 atom of lead and the 32 is equivalent to 2 atoms of oxygen (1 x 207 to 2 x 16), so the formula is simply PbO2
5. it is found that 54g of aluminium forms 150g of aluminium sulphide. Work out the formula of aluminium sulphide. (Relative atomic masses: Al = 27 and S = 32).
Amount of sulphur combined with the aluminium = 150 - 54 = 96gBy atomic ratio, the 54 of aluminium is equivalent to 2 atoms of aluminium and the 96 of sulphur is equivalent to 3 atoms of sulphur. Therefore the atomic ratio is 2 to 3, so the formula of aluminium sulphide is Al2S3
6. Magnesium + Oxygen ==> Magnesium oxide
2Mg + O2 ==> 2MgO
(atomic masses required: Mg=24 and O=16)
think of the ==> as an = sign, so the mass changes in the reaction are:
(2 x 24) + (2 x 16) = 2 x (24 + 16)48 + 32 = 2 x 40 and so 80 mass units of reactants = or produces 80 mass units of products (you can work with any mass units such as g, kg or tonne (1 tonne = 1000 kg)
7. iron + sulphur ==> iron sulphide (see the diagram at the top of the page!)Fe + S ==> FeS (atomic masses: Fe = 56, S = 32)If 59g of iron is heated with 32g of sulphur to form iron sulphide, how much iron is left unreacted? (assuming all the sulphur reacted)
From the atomic masses, 56g of Fe combines with 32g of S to give 88g FeS.This means 59 - 56 = 3g Fe unreacted.
8. When limestone (calcium carbonate) is strongly heated, it undergoes thermal decomposition to form lime (calcium oxide) and carbon dioxide gas.CaCO3 ==> CaO + CO2 (relative atomic masses: Ca = 40, C = 12 and O = 16)Calculate the mass of calcium oxide and the mass of carbon dioxide formed by decomposing 50 tonnes of calcium carbonate.
• (40 + 12 + 3x16) ==> (40 + 16) + (12 + 2x16)
• 100 ==> 56 + 44, scaling down by a factor of two,
• 50 ==> 28 + 22 so decomposing 50 tonnes of limestone produces 28 tonnes of lime and 22 tonnes of carbon dioxide gas.
9. he 'percent' % by mass composition of a compound in terms of its constituent elements is calculated as follows:
•  Calculate the formula or molecular mass of the compound
• Calculate the mass of the element in the compound, taking into account the number of atoms of the element in the compound formula
• Calculate (ii) as a percentage of (i)
10. Calculate the % of copper in copper sulphate, CuSO4
Relative atomic masses: Cu = 64, S = 32 and O = 16relative formula mass = 64 + 32 + 4x16 = 160only one copper atom of relative atomic mass 64% Cu = 64 x 100 / 160 = 40% copper by mass in the compound
11. Calculate the % of oxygen in aluminium sulphate, Al2(SO4)3
Relative atomic masses: Al = 27, S = 32 and O = 16relative formula mass = 2x27 + 3x(32 + 4x16) = 342there are 4 x 3 = 12 oxygen atoms, each of relative atomic mass 16, giving a total mass of oxygen in the formula of 12 x 16 = 192% O = 192 x 100 / 342 = 56.1% oxygen by mass in aluminium sulphate
12. alculate the % of water in hydrated magnesium sulphate MgSO4.7H2O
Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1relative formula mass = 24 + 32 + (4 x 16) + [7 x (1 + 1 + 16)] = 2467 x 18 = 126 is the mass of  waterso % water = 126 x 100 / 246 = 51.2 %
13. Molecular/formula mass =
total of all the atomic masses of all the atoms in the molecule/compound.
14. the diatomic molecules of the elements hydrogen H2 and chlorine Cl2 relative atomic masses, Ar: H = 1, Cl = 35.5
ormula masses, RMM or Mr, are H2 = 2 x 1 = 2, Cl2 = 2 x 35.5 = 71 respectively.
15. The compound sulphuric acid H2SO4relative atomic masses are H=1, S=32 and O=16
RMM or Mr = (1x2) + 32 + (4x16) = 98 (molecular mass of sulphuric acid)
16. alcium phosphate is also ionic but a more tricky formula to work out!(Ca2+)3(PO43-)2 or Ca3(PO4)3atomic masses: Ca = 40, P = 31, O =16
RFM or Mr = (3 x 40) + 3 x {31 + (4 x 16)} = (120) + (3 x 95) = 405
17. Atoms are
he smallest particles of matter whose properties we study in Chemistry. However from experiments done in the late 19th and early 20th century it was deduced that atoms were made up of three fundamental sub-atomic particles (listed below
18. write down the information about the subatomic particles in an atom 19. one mole of a substance is equal to
its Mr in grams
20. what is relative atomic mass
this is just a way of saying how heavy different atoms are compared with the mass of an atome of carbon-12
21. how do we calculate the Mr
its just all the relative atomic masses added together
22. how do we calculate the number of moles there are in a compund
• number of moles = mass in g (of that element or compound)
•                              ------------------------------------------
•                              Mr (of element or compound)
23. how do we calculate the percentage mass of an element in a compound
• Ar of that compound * number of atoms of that element
• --------------------------------------------------------- * 100
•  Mr of whole compound
24. how do we calculate the empirical formula
• list all the elements in the compound
• underneath them write their experimental masses of %'s
• divide each mass or % by the Ar of the element
• turn the numbers you get into a nice simple ratio
• get the ratio into its simplest form , and that tells you the emperical formula of the compund
25. percentage yield compares
actual and predicted yeild
26. the amount you get from a reaction is known as the
yeild
27. the more reactants you start with , the higher the actual yield will be . But the percentage yeild doesnt depend on
the amount of yeild you started with
28. percentage yeild =
• actual yeild (g)
• --------------       *100
• predicted yield (g)
29. percentage yeild is always somewhere between
0 and 100
30. even though atoms are never gained or lost , in real life
you never get a 100% percentage yeild . Some product or reactant always gets lost along the way - and that goes for industrial processes as well as school lab experiments
31. why do we never get a 100% percentage yeild
• the reaction may not go to completion because it is reversible
• or
• some of the produict may be lost when it is seperated from the reaction mixture
• or
• some of the reacts may react in differeny ways from the expected reaction
32. the relative atomic mass on an element (Ar) compares the mas of atoms of the element
with the C12 isotope . It is an average value for the isotopes of the element
33. isotope
Each of two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei, and hence differ in relative atomic mass but not in chemical properties; in particular, a radioactive form of an element
34. a compound was analysed and shown to have the following composition by mass and Mr of 135 dtermine the molecular formula
C = 14.8% , H = 3.5% , Cl = 43.6% , F = 31.8%
2CHClF
35. COP Chemicals makes substance X using a process that has a yield of 20% . The director at COP chemicals doesnt think that the lower % yeild is a problem because he is running at a very large profit . Suggest a reason why he should still increase the % yeild of a substance
a low yield means wasted chemicals which isnt sustainable . Increasing the yeild would save resources for the future
36. Aaliya calculated that they should produce a yield of 15g of barium sulphate . Howver after completing the experiment they found they only produced a yeild of 6g of barium sulphate calculate the % yield
40%
37. calculate how many hydrogen atoms are present in 0.04 moles of C2H6
0.24 H atoms
38. how many hydrogen atoms are there in 2.3g pg C2H5OH
0.3H atoms
39. iron oxide is reduced to iron inside a blast furnace using carbon . There were 3 stages involved
stage A = C + O2 = CO2
stage B = CO2 + C = 2CO
stage C = 3CO + Fe2O3 = 2Fe + 3CO2
if 10 g carbon are used in stage B , and all the carbon monoxide produced gets used up in stage C , what mass of CO2 is producd in stage C
73.392 g
40. what is the Ar of Na
23
41. what is the Ar of Mg
24
42. what is the Ar of K
39
43. what is the Ar of Ca
40
44. what is the Ar of Ni
59
45. what is the Ar of Cu
63.5
46. what is the Ar of Zn
65
47. what is the Ar of Li
7
48. what is the Ar of Al
27
49. what is the Ar of C
12
50. what is the Ar of N
14
51. what is the Ar of O
16
52. What is the Ar of F
19
53. what is the Ar of Ne
20
54. what is the Ar of He
4
55. what is the Ar of S
32
56. what is the Ar of Cl
35.5
57. what is the Ar of Ar
40
 Author: ghoran ID: 168122 Card Set: GCSE Chemistry Updated: 2012-08-30 14:19:19 Tags: quantitative chemistry Folders: Description: revision Show Answers: