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1. Convert the following numbers to scientific notation:
a. 124
1.24 x 102
b. 0.004852
4.852 x 10-3
c. 54.21
5.421 x 101
d. 0.000084
8.4 x 10-5
2. Write the following numbers in standard notation
a. 1.33 x 10-3
0.00133
b. 5.77 x 105
577,000
3. Round the following numbers to 3 sig figs
a. 19.9994
20.0
b. 8,439
8,440
8
c. 5.3861 x 10
5.39 x 108
4. How many atoms are in 3.4 mol Cu?
3.4 (6.022 x 1023) = 2.0 x 1024 atoms of Cu
5. Calculate the mass of 1.28 mol of Fe
1.28 x 55.845 = 71.5 g of Fe
6. How many moles of atoms are in 4.6 x 1024 atoms of Pb?
4.6 ´10 24
= 7.6 moles of Pb
6.022 ´10 23
7. 285.33 g of NaNO3 represent how many moles?
285.33
= 3.3571 moles of NaNO3
84.994
8. How many atoms are in 9.7 x 10-3 mol C?
(9.7 x 10-3) (6.022 x 1023) = 5.8 x 1021 atoms of C
9. Determine the mass of 2.48 x 1022 atoms of HCl
2.48 ´10 22
= 0.0411823 mol of HCl
6.022 ´10 23
(0.0411823) (36.4609) = 1.50 g of HCl
10. Calculate the mass percent composition for each of the following compounds.
a. C2H2
2(12.011)
(100) = 92.258% C
2(12.011) + 2(1.0079)
2(1.0079)
(100)
2(12.011) + 2(1.0079)
= 7.7418% H
b. C3H6
3(12.011)
(100) = 85.629% C
3(12.011) + 6(1.0079)
6(1.0079)
(100)
3(12.011) + 6(1.0079)
= 15.230% H
11. Calculate the empirical formula for the compound given the following masses
of their constituent elements.
1.245 g Ni, 5.381 g I
1.245
5.381
= 0.02121 mol Ni
= 0.04240 mol I
58.693
126.90
0.02121
=
0.02121 1
NI2
0.04240
=
0.02121 1.999
12. Calculate the empirical formula for pineapple oil, Ethyl butyrate:
C – 62.04%, H – 10.41%, O – 27.55%
62.04
10.41
27.55
= 5.1657 mol C
= 10.328 mol H
= 1.722 mol O
12.011
1.0079
15.999
5.16757
10.328
= 3 mol C
= 6 mol H
1 mol O
1.722
1.722
C3H6O
13. The following are the molar masses and empirical formulas of several
compounds. Find the molecular formula of each compound.
a. 284.77 g/mol – CCl
CCl = 12.011 + 35.453 = 47.464 g/mol
284.77
=6
47.464
so
C6Cl6
b. 131.39 g/mol – C2HCl3
C2HCl3 = 2(12.011) + 1.0079 + 3(35.453) = 131.39 g/mol
Molar mass is the same as empirical mass so the EF = MF
C2HCl3