Solved: Determine the chemical formula of each compound and refer to the formula to

Question

Problem 109P

Determine the chemical formula of each compound and refer to the formula to calculate the mass percent composition of each constituent element.

(a) copper(II) iodide

(b) sodium nitrate

(c) lead(II) sulfate

(d) calcium fluoride

Answer 1

Step-by-Step Solution:

Step 1 of 3

Solution 109P

Here, we have to calculate the mass percent composition of each constituent element.The mass percent composition or mass percent of an element is that the percentage of the element present in compound’s total mass.

% mass = (mass of element in 1 mole of the compound) / (molar mass of the compound) x 100%

(a) copper(II) iodide

The chemical formula of the compound is CuI2.

Molar mass of Copper iodide = 63.546 g/mol + 2x 126.904 g/mol

= 317.354 g/mol

Molar mass of Cu = 63.456 g/mol

Molar mass of I = 126.904 g/mol

Thus,

Mass percent of I = $\frac{2x\ 126.904\ g/mol}{317.354\ g/mol}$ x 100

= 79.97% $\sim{}\ 80\%\$

Mass percent of Cu = $\frac{63.456g/mol}{317.354\ g/mol}$ x 100

= 19.99 % $\sim{}20\%$

Thus, the chemical formula of the compound is CuI2.

And the mass percent of Cu is 19.99 % $\sim{}20\%$ and

The mass percent of I is 79.97% $\sim{}\ 80\%\$

(b) sodium nitrate

The chemical formula of the compound is NaNO3.

Molar mass of Na = 22.989 g/mol

Molar mass of N = 14.0 g/mol

Molar mass of O = 16.0 g/mol

Molar mass of NaNO3 = 22.989 g/mol + 14.0 g/mol +3 x 16.0 g/mol

= 84.989 g/mol

Thus,

Mass percent of Na = $\frac{22.989\ g/mol}{84.989\ \ g/mol}$ x 100

= 27.05 %

Mass percent of N = $\frac{14.0\ g/mol}{84.989\ \ g/mol}$ x 100

= 16.47 %

Mass percent of O = $\frac{3\ x16.0\ g/mol}{84.989\ \ g/mol}$ x 100

= 56.47 %

Thus, the chemical formula of the compound is NaNO3 and

the mass percent of Na is 27.05%

The mass percent of N is 16.47 %

The mass percent of O is 56.47 %

(c) lead(II) sulfate

The chemical formula of the compound is PbSO4.

Molar mass of Pb = 207.2 g/mol

Molar mass of S = 32.065 g/mol

Molar mass of O = 16.0 g/mol

Molar mass of PbSO4 = 207.2 g/mol + 32.065 g/mol + 4 x 16.0 g/mol

= 303.265 g/mol

Thus,

Mass percent of Pb = $\frac{207.2\ g/mol}{303.265\ \ g/mol}$ x 100

= 68.32 %

Mass percent of S = $\frac{32.065\ \ g/mol}{303.265\ \ g/mol}$ x 100

= 10.57 %

Mass percent of O = $\frac{4\ x16.0\ g/mol}{303.265\ \ g/mol}$ x 100

= 21.10 %

Thus, the chemical formula of the compound is PbSO4 and

the mass percent of Pb is 68.32 %

The mass percent of S is 10.57 %

The mass percent of O is 21.10 %

(d) calcium fluoride

The chemical formula of the compound is CaF2.

Molar mass of Ca =40.078 g/mol

Molar mass of F = 18.998 g/mol

Molar mass of CaF2 = 40.078 g/mol + 2x 18.998 g/mol

= 78.074 g/mol

Thus,

Mass percent of Ca = $\frac{40.078\ \ g/mol}{78.074\ \ g/mol}$ x 100

= 51.333 %

Mass percent of F = $\frac{2x\ 18.998\ \ g/mol}{78.074\ \ g/mol}$ x 100

= 48.666 %

Thus, the chemical formula of the compound is CaF2 and

the mass percent of Ca is 51.33 %

The mass percent of F is 48.66 %

Step 2 of 3

Chapter 6, Problem 109P is Solved

Step 3 of 3

Solved: Determine the chemical formula of each compound and refer to the formula to

Solution for problem 109P Chapter 6

Chapter 6, Problem 109P is Solved

Textbook: Introductory Chemistry

Edition: 5

Author: Nivaldo J Tro

ISBN: 9780321910295

Other solutions

People also purchased

Related chapters