Page 55 - Theory and Problems of BEGINNING CHEMISTRY

44 ATOMS AND ATOMIC MASSES [CHAP. 3

EXAMPLE 3.16. Arrange the 12 electrons of magnesium into shells.
Ans. 2 8 2

The first two electrons fill the first shell, and the next eight fill the second shell. That leaves two electrons in the third
shell.

The number of outermost electrons is crucial to the chemical bonding of the atom. (See Chap. 5.) For main
group elements, the number of outermost electrons is equal to the classical group number, except that it is 2 for
helium and 8 for the other noble gases. (It is equal to the modern group number minus 10 except for helium and
the first two groups.)
Electron dot diagrams use the chemical symbol to represent the nucleus plus the inner electrons and a dot
to represent each valence electron. Such diagrams will be extremely useful in Chap. 5.

EXAMPLE 3.17. Write the electron dot symbols for elements Na through Ar.
Ans. Na Mg Al Si P S Cl Ar

Solved Problems

ATOMIC THEORY
3.1. If 10.0 g of sodium and 20.0 g of chlorine are mixed, they react to form 25.4 g of sodium chloride.
Calculate the mass of chlorine that does not react.
Ans. The Law of Conservation of Mass allows us to determine the mass of unreacted chlorine.
10.0g + 20.0g = 25.4g + x

x = 4.6g

3.2. A sample of a compound contains 1.00 g of carbon and 1.33 g of oxygen. How much oxygen would be
combined with 7.00 g of carbon in another sample of the same compound? What law allows you to do
this calculation?

Ans. Since it is the same compound, it must have the same ratio of masses of its elements, according to the Law
1.33gO
of Definite Proportions. Thus there is 7.00gC = 9.31gOinthe second sample.
1.00gC
3.3. Determine in each part whether the two samples are samples of the same compound or not. What law
allows this calculation?

Sample 1 Sample 2
Element 1 Element 2 Element 1 Element 2

(a) 6.11 g 8.34 g 7.48 g 10.2 g
(b) 43.1 g 88.3 g 12.7 g 17.3 g

Ans. If they are the same compound, they will have the same ratio of masses according to the Law of Definite
Proportions.
(a) 6.11 g element 1 7.48 g element 1
= 0.733 = 0.733
8.34 g element 2 10.2 g element 2
These are samples of the same compound.
(b) 43.1 g element 1 12.7 g element 1
= 0.488 = 0.734
88.3 g element 2 17.3 g element 2
These are not samples of the same compound.