Stoichiometry Chemistry Worksheets With Answers Page 10
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111. Hydrogen’s atomic weight = 1, chlorine’s atomic weight = 35, so hydrogen makes up 1/36, or
2.7%, of the compound.
2. Hydrogen’s atomic weight = 1, oxygen’s atomic weight = 16; there are two hydrogens here, so
it makes up 2/18 of the total weight, or 11%.
3. Hydrogen’s atomic weight = 1, phosphorus’s weight = 31, oxygen’s weight = 16, so after you
add up the total weight to get 3(1) + 31 + 4(16) = 98, hydrogen makes up 3/98 = 3%.
4. Hydrogen’s weight = 1, sulfur’s = 32, and oxygen’s = 16. The total weight is equal to 2(1) + 32
+ 4(16) = 98, and hydrogen makes up 2% of this.
5. Hydrogen’s weight = 1, fluorine’s weight = 19, so hydrogen makes up 1/19, or 5.2%, of the
molecule by weight.
As you can see, choice B, in which hydrogen makes up 11% of the total compound, is the correct answer.
6.
F, T
(Do not fill in CE.) The first statement is false—in most chemical reactions, most times the number of moles of the
compounds involved is not equal. The second statement, however, is true. Once a limiting reagent has been consumed in
the course of a reaction, the reaction can no longer proceed.
7.
A
To solve this problem, first turn the percentages into grams: C becomes 96 g carbon, so the remaining 4% is hydrogen, so
there are 4 g of hydrogen present. Now convert these masses into moles: the moles of C = 96 g (1 mol/12 g/mol) = 8 mol.
For hydrogen, H = 4 g (1 mol/1 g/mol) = 4 mol. The C:H ratio is 8:4; remember that the empirical formula is the formula that
shows the relative numbers of the kinds of atoms in a molecule, so this simplifies to 2:1, and the empirical formula is C
H.
2
8.
E
Turn the percentages into grams and find the number of moles of each element. Since the compound is 48% C and 4% H, it
must also be 48% O to make the percentages total 100%. Simplify the mole:mole ratio to get the empirical formula.
Calculate the empirical molar mass. The molecular mass will be some multiple of the empirical mass. Mol C = 48 g (1 mol/12
g/mol) = 4 mol. Mol H = 4 g (1 mol/1 g/mol) = 4 mol. Mol O = 48 g (1 mol/16 g/mol) = 3 mol. The empirical formula is then
C
H
O
. The empirical molar mass = 4(12) + 4(1) + 3(16) = 100, so the molecular formula is twice the empirical formula, or
4
4
3
C
H
O
.
8
8
6
9.
B
This is not a limiting reactant problem since only one amount is given here, 7.0 g of ethene. First find the number of moles of
the substance, in this case ethene: 7 g (1 mol/28 g/mol) = 0.25 mol, then use mole:mole to determine the moles of the rest
of the compounds involved in the reaction. To find the grams of CO
, you would do the following: 1 mol
44 g/mol = 44 g
2
CO
.
2
Molar mass
28
32
44
18
C
H
+
2CO
+
2H
O
Balanced equation
2
4
2
2
3O
2
0.25
1.25
0.5
0.5
No. of moles
7 g
22 g
Amount
10.
D
Here, determine the moles of methane and use the mole:mole ratio to determine the number of moles, then grams of CF
4
that can be produced.
Molar mass
16
38
88
20
Balanced equation
CH
+
CF
4HF
4
4
4F
4
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