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Example #1: An element X has a dibromide with the empirical formula XBr2 and a dichloride with the empirical formula XCl2. The dibromide is converted to the dichloride according to the equation:
XBr2 + Cl2 --> XCl2 + Br2
If the complete conversion of 1.710 g of XBr2 results in the formation of 1.014 g of XCl2, what is the atomic mass of the element X? What chemical element is X?
1) We know from the balanced equation that one mole of XBr2 will produce one mole of XCl2.
moles of XBr2 that react = 1.710 / (M + 159.808) <--- where M is the atomic weight of element X
moles of XCl2 that form = 1.014 / (M + 70.906) <--- where M is the atomic weight of element X
2) The moles of XBr2 that react are equal to the moles of XCl2 that are produced:
1.710 1.014 ––––––––––– = ––––––––––– (M + 159.808) (M + 70.906)
(1.710) (M + 70.906) = (1.014) (M + 159.808)
1.710M + 121.24926 = 1.014M + 162.045312
0.696M = 40.796052
M = 58.62 g/mol (to 4 sig figs)
3) The element, most probably is nickel, with an atomic weight of 58.69.
Example #2: A 3.41 g sample of a metallic element (M) reacts with 0.0158 mol of a gas (X2) to form 4.52 g MX. What are the identities of M and X?
M + 1⁄2X2 ---> MX
mass of X2 ---> 4.52 − 3.41 = 1.11 g
molecular weight of X2 ---> 1.11 g / 0.0158 mol = 70.3 g/mol
X2 is chlorine, whose molar mass is 2(35.45) = 70.9 g/mol
Use mole-mole stoichiometry to determine moles of M that react:
1 mol M x ––––––––– = –––––––––––– 0.5 mol X2 0.0158 mol X2
x = 0.0316 moles M
3.41 g / 0.0316 mol = 107.9 g/mol
M is silver
Example #3: For the reaction below, when 0.5000 g of XI3 reacts completely, 0.2360 g of XCl3 is obtained. Calculate the atomic weight of element X and identify it.
2XI3 + 3Cl2 ---> 2XCl3 + 3I2
1) From the balanced equation, we know that the moles XI3 used equals moles XCl3 produced. Therefore:
0.5000 0.2360 ––––––––– = –––––––––––– (x + 381 g/mol) (x + 106.5 g/mol)
The 381 is the weight of three iodines and the 106.5 is the weight of three chlorines.
2) Solving for x, we find it equal to 138.9. This is the atomic weight of lanthanum.
Example #4: If 0.520 grams of XCl3 are treated with iodine, 0.979 g of XI3 are produced. What is the chemical symbol for the element X?
2XCl3 + 3I2 ---> 2XI3 + 3Cl2
1) From the balanced equation, we know that the moles XCl3 used equals moles XI3 produced. Therefore:
0.520 g / (x + 106.5 g/mol) = 0.979 g / (x + 381 g/mol)
The 106.5 is the weight of three chlorines and the 381 is the weight of three iodines.
2) Solving for x, we find it equal to 204.5. This is the atomic weight of thallium and its symbol is Tl.
Example #5: Consider the reaction involving unknown element X:
F2 + 2XBr ---> Br2 + XF
When 5.500g of XBr reacts, 3.693g of Br2 is produced. Identify element X.
1) Moles of Br2:
3.693 g / 159.808 g/mol = 0.023109 mol
2) Moles of XBr:
0.023109 mol times 2 = 0.046218 mol
3) Molar mass of XBr:
5.500 g / 0.046218 mol = 119.0 g/mol
4) Identity of X
119.0 − 79.904 = 39.0
X is potassium
Example #6: 0.126 g of a metal, M, reacts with HCl to form hydrogen gas and MCl3. It is found that, upon complete reaction of M, 0.01411 g of hydrogen gas was formed. Calculate the atomic mass of M and identify what element it most probably is.
1) Let us write a balanced chemical equation for the reaction:
2M + 6HCl ---> 2MCl3 + 3H2
2) Determine moles of hydrogen gas produced:
0.01411 g / 2.016 gmol = 0.006999 mol
3) Determine moles of M that was consumed (use the M:H2 molar ratio):
2 x ––– = ––––––––––– 3 0.006999 mol
x = 0.004666 mol
4) Determine molar mass of M:
0.126 g / 0.004666 mol = 27.0 g/mol
M is aluminum.
Example #7: 2.56823 grams of sodium chloride furnishes 6.2971 grams of silver chloride in the following reaction:
NaCl + Ag ---> AgCl + Na
Calculate the atomic weight of sodium.
1) Determine moles of AgCl:
6.2971 g / 143.32 g/mol = 0.043937343 mol
2) NaCl and AgCl are in a 1:1 molar ratio. Therefore, 0.043937343 mol of NaCl was consumed.
3) Determine the molecular weight (MW) of NaCl:
2.56823 g / MW = 0.043937343 mol
MW = 58.45 g/mol
4) Knowing the atomic weight of chlorine to be 35.45 g/mol, we determine the weight of sodium:
58.45 − 35.45 = 23.00 g/mol
Example #8: Silver phosphate is found by careful analysis to contain 77.300 per cent silver. What is the calculated atomic weight of phosphorus.
1) The formula for silver phosphate is Ag3PO4.
2) Assume 100. g of silver phosphate is present. Therefore:
77.300 g is silver
22.700 g is phosphate
3) Moles of silver present:
77.300 g / 107.868 g/mol = 0.7166166 mol
4) The molar ratio between silver and the phosphate group is 3:1. Therefore, moles of the phosphate group present is 0.2388722 mol (which is 0.7166166 divided by 3).
5) Determine atomic weight of phosphorous:
22.700 g ––––––––– = 0.2388722 mol (X + 63.9976)
0.2388722X + 15.2872475 = 22.700
0.2388722X = 7.4127525
X = 31.0 g/mol
By the way, 63.9976 is the weight of 4 oxygens.
Example #9: By experiment, it was found that the mass ratio of silver bromide to silver chloride to be 1.310171. What is the atomic weight of bromine?
1) Here is the chemical reaction this question is based on:
AgBr + 1⁄2Cl2 ---> AgCl + 1⁄2Br2
The key point will be the 1:1 molar ratio between AgBr and AgCl.
2) The mass ratio is 1.310171 to 1. The molar ratio is known to be 1 to 1. This expression can then be written:
1.310171 (107.868 + X) <--- molecular weight of AgBr ––––––– = ––––––––––––––– 1 (107.868 + 35.453) <--- molecular weight of AgCl
Due to the 1:1 molar ratio, we know that 1.310171 g of AgBr contains the same number of moles as does 1 g of AgCl. Therefore, the mass ratio from the experiment is equal to the mass ratio of the molecular weights.
3) We solve for X, the atomic weight of bromine:
107.868 + X = (1.310171) (143.321)
107.868 + X =187.775
X = 79.9 g/mol
4) Note the different set up in example #10.
Example #10: What is the atomic weight of copper if 1.0000 gram of Cu is obtained by the reduction of 1.2518 grams of CuO?
1) This is the reaction used:
CuO ---> Cu + 1⁄2O2
The key point is that the CuO to Cu molar ratio is 1:1
2) A way to state that 1:1 molar ratio is this:
moles of CuO consumed = moles of Cu produced
1.2518 1.0000 ––––––––– = ––––––– (X + 16.00) X
1.2518X = X + 16.00
0.2518X = 16.00
X = 63.54 g/mol
4) Note the different set up in example #9.
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