Review your understanding /33 |
1) Non metal X can be identified from its relative isotopic masses. The table below shows the abundances of the isotopes:
Isotope | Isotope 1 | Isotope 2 |
Relative isotopic mass | 37 | 35 |
% abundance | 25.0 | 75.0 |
a) Define relative atomic mass
The weighted mean mass of an atom of an element [1] compared with 1/12th of the mass of of an atom of carbon - 12 [1]
b) Calculate the relative atomic mass of a sample of X. Give your answer to 3 significant figures. 35.5 [1] answer [1] 3sf
c) Use your answer to (b) and the data sheet to identify element X Cl [1]
2) A sample of gas was found to contain 1.630g of carbon and 0.272g of hydrogen by mass and an Mr of 28.
Calculate the molecular formula of this compound.
C | H | ||
Mass | 1.630 | 0.272 | |
Moles | 1.630 / 12 | 0.272 / 1 | |
0.136 [1] | : | 0.272 [1] | |
divide by the smallest | 0.136 / 0.136 | 0.272 / 0.136 | |
1 | : | 2 |
Empirical formula = CH2 [1] Mr of Ef = 14. Mr of molecule = 28. Therefore molecular formula = C2H4 [1]
3) A student added 2.74g of Barium to water according to the following reaction:
Ba(s) + 2H2O(l) à Ba(OH)2(aq) + H2(g)
The resulting solution was then made up to 250cm3 with distilled water.
a) Calculate how many moles of Ba reacted
Moles = mass / Ar 2.74 / 137.3 = 0.0199563 [1] = 0.0200 3sf [1]
b) Calculate the volume of hydrogen that would be produced assuming 1 mole of gas occupies 24dm3
Molar ratio = 1:1, so 0.0200 moles H2 made [1] Vol = Moles x 24 = 0.48dm3 [1]
c) Calculate the concentration of the Ba(OH)2
Moles of Ba(OH)2 = 0.0200 (1:1) ratio [1] Conc = moles / vol, 0.0200 / 0.25 = 0.0800 mol dm-3 [1]
d) Using oxidation numbers identify which element has been oxidised.
Ba [1] as its oxidation number has gone from 0 to 2+ [1] showing electrons have been lost [1]
4) A student added magnesium oxide to 25cm3 of 2.00 mol dm-3 nitric acid until all the nitric acid reacted.
a) Write a balanced chemical equation for the above reaction, include state symbols.
MgO(s) + 2HNO3(aq) à Mg(NO3)2(aq) + H2O(l) [1] All formulas [1] Balanced [1] state symbols
b) How would the student know when the reaction was complete?
No more MgO would react / dissolve away [1]
c) Calculate how many mole of nitric acid was used in the reaction.
Moles = C x V, 2.00 x 0.025 = 0.0500 moles [1] answer [1] 3sf
d) Deduce the number of moles of magnesium oxide that reacted with this amount of nitric acid.
2 : 1 ratio, therefore 0.0500 / 2 = 0.0250 moles
e) Calculate the mass of magnesium oxide reacted with the nitric acid, give your answer to the nearest whole number [3]
Mr = 40.3 [1] Mass = moles x Mr, 0.025 x 40.3 = 1g [1] whole number [1]
5) A 2.95g sample of the hydrated crystal CoI2.nH2O was heated until there was no further change in mass.
The final mass of the anhydrous crystals was 2.190g.
a) Calculate n and give the formula of the salt.
CoI2 | H2O | ||
Mass | 2.190 | (2.950 - 2.190 =) 0.760 | |
Moles | 2.190 / 312.7 | 0.760 / 18 | |
0.00700 [1] | : | 0.0422 [1] | |
divide by the smallest | 0.00700 / 0.00700 | 0.0422 / 0.00700 | |
1 | : | 6 |
CoI2.6H2O [1]
b) Write an equation for the change that took place. Include state symbols in your answer.
CoI2.nH2O(s) à CoI2(s) + 6H2O(g) [1] Balanced equation [1] state symbols