E) Nicotine, contains carbon, hydrogen and nitrogen. When a 5.00-g sample of nic
ID: 544868 • Letter: E
Question
E) Nicotine, contains carbon, hydrogen and nitrogen. When a 5.00-g sample of nicotine was combusted in excess of oxygen, 13.56 g of CO2, 3.886 g of H2O and some nitrogen oxides were produced. The empirical formula for this compound is A) CsH7N B) C4H3N C) C H4N D) C3HsN2 E) C2H7N Calcium carbonate decomposes on heating to produce calcium oxide and carbon dioxide gas as follows: CaO(s) CO2(g) CaCO3(s) heat + + If the yield of the reaction is only 73.0%, what is the minimum amount of calcium carbonate needed to produce 12.7 g of calcium oxide? A) 31.1 g B) 17.4 g C) 76.8 g D) 29.3 g E) 73.1 g OH Which are the spectator ions in the following reaction? (Assume both H:SO, and K dissociates completely in aqueous solution)Explanation / Answer
Q11
Empirical formula is the least coefficient formula, that is
CxHyNz
we must find x,y,z via gravimetry.
Typically we do this relating to moles of C,H,N
mol of CO2 = mass of CO2/MW of CO2 = (13.56)/(44) = 0.30818 mol of CO2
1 mol of CO2 = 1 mol of C ---> 0.30818 mol of CO2 = 0.30818 mol of C
mol of H2O = mass of H2O/MW of H2O= (3.886)/(18) = 0.2158 mol of H2O
1 mol of H2O= 2 mol of H ---> 0.2158 mol of H2O= 0.2158 *2 = 0.4316 mol of H
Now... mol of Nitrogen:
Mass of N = Mass of sample - Mass of C - Mass of H
Mass of C = mol of C * MW of C = 0.30818 *12 = 3.69816 g
Mass of H = mol of H * MW of H = 0.4316 *1= 0.4316 g
Mass of N = 5- 3.69816 -0.4316 = 0.87024 g of N
mol of N = mass of N / MW of N = (0.87024)/(14) = 0.06216 mol of N
Ratios:
C:N = 0.30818 / 0.06216 = 5
H:N = 0.4316 / 0.06216 = 7
H:C = 0.4316 /0.30818 = 1.4 --> 7/5
then..
C5H7N
is the empriical formula
Q12.
mol of CaO = mass/MW = 12.7/56.0774 = 0.22647
yield is only 73% so
mol of CaCO3 required = 0.22647 for 100%
0.22647 / 0.73 --> 0.31023 mol of CaCO3 required
mass = mol*MW = 0.31023*100 = 31.02 g
choose A
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