152
Stoichiometry I: Calculations Based on Formulas
structures and reactivities are the same, their masses actually may vary. These
different mass forms are called isotopes. It is because the percentage of each
isotope of an element is always the same throughout nature that the mass of
each atom of that element appears to be the same for all. The mass spectrometer provides a means of accurately finding the percentage of each isotope and its
actual mass. For carbon, for example, two isotopes are found: 98.892% of one
isotope whose mass is 12.00000 d/atom, and 1.108% of the other whose mass is
13.00335 d/atom. If we should take a million carbon atoms at random, then
988,920 of them would each weigh 12.00000 d and 11,080 of them would each
weigh 13.00335 d. The total weight would be
(988,920 atoms)( 12.00000 d/atom) = 11,867,040 d
(11,080 atomsK 13.00335 d/atom) =
144,080 d
Total weight of 10
6 atoms = 12,011,120 d
Average weight of one atom = —f^—7
= 12.01112
atoms
'
atom
This average weight is the weight that«// carbon atoms appear to have as they
are dealt with by chemists. The average atomic weights for all the elements have
been determined in a similar way, and it is these averages that are listed in the
relative atomic weight scale inside the back cover.
PROBLEMS A
1 Find the percentage composition of (the percentage by weight of each element
in) each of the following compounds
(a) N 2 O
(g) Na^Og 5H 2 O
(b) NO
(h) Ca(CN) 2
(c) NO 2
(i) (NH^CO,
(d) Na 2 SO 4
(j) UO 2 (N00 2 6H 2 O
(e) Na 2 S 2 O 3
(k) Penicillin, C 16 H 26 O 4 N 2 S
(f) Na2SO 4 10H 2 O
2 What is the weight of 1 00 mole of each compound in Problem 1 '
3 How many moles are in 1 00 Ib of each compound in Problem I '
4 Find the number of molecules in
(a) 25 0 g H 2 0
(b) 1.00 oz of sugar, C 12 H 22 O U
(c) 1 00 microgram of NH-,
(d) 5 00 ml of CC1 4 whose density is 1 s94 g/ml
NOTF Problems concerning the determination of appioximate atomic weights by the rule of
Dulong and Petit may be found in Chapter 14
Stoichiometry I: Calculations Based on Formulas
structures and reactivities are the same, their masses actually may vary. These
different mass forms are called isotopes. It is because the percentage of each
isotope of an element is always the same throughout nature that the mass of
each atom of that element appears to be the same for all. The mass spectrometer provides a means of accurately finding the percentage of each isotope and its
actual mass. For carbon, for example, two isotopes are found: 98.892% of one
isotope whose mass is 12.00000 d/atom, and 1.108% of the other whose mass is
13.00335 d/atom. If we should take a million carbon atoms at random, then
988,920 of them would each weigh 12.00000 d and 11,080 of them would each
weigh 13.00335 d. The total weight would be
(988,920 atoms)( 12.00000 d/atom) = 11,867,040 d
(11,080 atomsK 13.00335 d/atom) =
144,080 d
Total weight of 10
6 atoms = 12,011,120 d
Average weight of one atom = —f^—7
= 12.01112
atoms
'
atom
This average weight is the weight that«// carbon atoms appear to have as they
are dealt with by chemists. The average atomic weights for all the elements have
been determined in a similar way, and it is these averages that are listed in the
relative atomic weight scale inside the back cover.
PROBLEMS A
1 Find the percentage composition of (the percentage by weight of each element
in) each of the following compounds
(a) N 2 O
(g) Na^Og 5H 2 O
(b) NO
(h) Ca(CN) 2
(c) NO 2
(i) (NH^CO,
(d) Na 2 SO 4
(j) UO 2 (N00 2 6H 2 O
(e) Na 2 S 2 O 3
(k) Penicillin, C 16 H 26 O 4 N 2 S
(f) Na2SO 4 10H 2 O
2 What is the weight of 1 00 mole of each compound in Problem 1 '
3 How many moles are in 1 00 Ib of each compound in Problem I '
4 Find the number of molecules in
(a) 25 0 g H 2 0
(b) 1.00 oz of sugar, C 12 H 22 O U
(c) 1 00 microgram of NH-,
(d) 5 00 ml of CC1 4 whose density is 1 s94 g/ml
NOTF Problems concerning the determination of appioximate atomic weights by the rule of
Dulong and Petit may be found in Chapter 14