ChemTeam: Determine identity of an element from a binary formula and a percent composition

Determine identity of an element from a binary formula and a percent composition
Ten Examples

Example #1: A metal (M) forms an oxide with the formula MO. If the oxide contains 39.70% O by mass, what is the identity of M?

Solution:

1) Assme 100 g of the compound is present:

39.70 g is O; 60.30 g is M

2) Calculate moles of oxygen:

39.70 g / 16.00 g/mol = 2.48 mol

3) Determine moles of M:

The M to O molar ratio is 1:1
2.48 is to 1 as x is to 1
x = 2.48 mol of M

4) Calculate atomic weight of M and identify it:

60.30 g / 2.48 mol = 24.314 g/mol
M is magnesium

Example #2: A certain metal sulfide, MS₂, is determined to be 40.064% sulfur by mass. What is the identity of the metal M?

Solution #1:

1) Assume 100 g of the compound is present. Therefore:

40.064 g is sulfur; 59.936 g is M

2) Calculate moles of sulfur:

40.064 g / 32.065 g/mol = 1.249462 mol

3) Use molar ratio to get moles of M:

the M to S molar ratio is 1 : 2
x is to 1 as 1.249462 mol is to 2
x = 0.624731 mol of M

4) Determine atomic weight of M:

59.936 g / 0.624731 mol = 95.94 g/mol
M is molybdenum

Solution #2:

1) Determine molar mass of MS₂

64.130 is to 0.40064 as x is to 1
x = 160.069 g/mol

2) Subtract weight of sulfur:

160.069 − 64.130 = 95.94 g/mol

N.B. the 64.130 value is the weight of two sulfurs.

Example #3: A compound is found to have the formula XBr₂, in which X is an unknown element. Bromine is found to be 71.55% of the compound. Determine the identity of X.

Solution #1:

1) Determine molar mass of XBr₂

159.808 is to 0.7155 as x is to 1
x = 223.3515 g/mol

2) Subtract weight of the two bromines:

223.3515 − 159.808 = 63.543 g/mol
The element is copper.

Solution #2:

Let us assume 100 g of the compound is present. That means 71.55 g of Br is in the compound.
71.55 g / 79.904 g = 0.89545 mol
0.89545 mol is to 2 as y is to 1
y = 0.447725 mole of X is present in 100 g of XBr₂
100 g − 71.55 g = 28.45 g of X in 100 g of XBr₂
28.45 g / 0.447725 = 63.54 g/mol
X is copper

Example #4: A compound whose empirical formula is XF₃ consists of 64.79% F by mass. What is the atomic mass of X? What element is it?

Solution:

Let us assume 100 g of XF₃ is present.
64.79 g of it is F and 35.21 g of it is X
64.79 g / 18.9984 g/mol = 3.4103 mol
The molar ratio of X to F is 1 to 3
1 is to 3 as y is to 3.4103 mol
y = 1.13677 mol (this is how many moles of X are in the 100 g of XF₃)
35.21 g / 1.13677 mol = 30.97 g/mol
X is phosphorous.

Example #5: An element X forms the pentabromide XBr₅. Analysis of XBr₅ shows that it contains 92.81% Br by mass. What is element X?

Solution:

1) Let us assume 100 g of XBr₅ is present. That means this:

92.81 g of Br is present and 7.19 g of X

2) Calculate moles of Br:

92.81 g / 79.9 g/mol = 1.1616 mol

3) Determine moles of X:

1.1616 mol is to 5 as x is to 1
x = 0.23232 mol

4) Determine atomic weight of X:

7.19 g / 0.23232 mol = 30.95 g/mol
The element is phosphorous.

Example #6: An element X forms the tetrachloride XCl₄. Analysis of XCl₄ shows that it contains 40.63% Cl by mass. Identify X.

Solution:

1) Assume 100 g of the compound is present. This means:

40.63 g of chloride in the 100 g and 59.37 g of X

2) Determine moles of chloride:

40.63 g / 35.453 g/mol = 1.146 mol

3) Determine moles of X:

1.146 mol is to 4 as x is 1
x = 0.2865 mol of X is present

4) Determine atomic weight of X:

59.37 g / 0.2865 mol = 207.2 g/mol
X is lead, PbCl₄

Example #7: A metal M forms an oxide M₂O₃ that contains 68.4% of the metal by mass. Calculate the atomic weight of the metal.

Solution:

68.4 is to 2x as 31.6 is to 48
The 48 comes from 16 times 3 because there are 3 oxygens. It is 2x because there are two atoms of M.
x = 51.95, the atomic weight of chromium.

Example #8: A certain metal hydroxide, M(OH)₂, contains 32.80% oxygen by mass. What is the identity of the metal M?

Solution #1:

1) Write an expression for the molar mass of the compound:

Let M = the molar mass of the metal.
M + (2 x 16.00) + (2 x 1.01)
M + 34.02

2) Write an expression for the given mass percent of oxygen:

[32.00 / (M + 34.02)] x 100% = 32.8%
32.00 / (M + 34.02) = 0.328

3) Algebra!

32.00 = (0.3280)(M + 34.02)
32.00 = 0.3280M + 11.16
20.84 = 0.3280M
M = 63.54
Copper.

Solution #2:

1) Determine mass percent of hydrogen:

32.80 is to 32.00 as x is to 2.02
x = 2.07%

2) Determine mass percent of M:

100% − (32.80 + 2.07) = 65.13%

3) Determine atomic weight of M:

ONE metal atom has (65.13/32.80) times the weight of TWO oxygen atoms.
the mass of the metal atom = (65.13/32.80) (2 x 16.00) = 63.54 amu
Copper.

Solution #3:

1) Determine the molar mass of M(OH)₂:

32.00 is to 0.3280 as x is to 1
x = 97.56 g/mol

2) Subtract the weight of two hydroxides:

97.56 − 34.02 = 63.54 g/mol
Copper.

Solution #4:

1) Assume 100.0 g of the compound is present. Therefore, the mass of oxygen in the compound:

(100.0 g) (0.328) = 32.8 g

2) Moles of O in the compound:

32.8 g / 16.00 g/mole = 2.05 mole

3) Mass of H in the compound:

(2.05 mol ) (1.01 g/mol) = 2.07 g

4) Mass of metal in the 100 g of compound:

100.0 g − 32.8 g − 2.07 g = 65.13 g

5) From the formula:

moles of M = moles of O / 2 = 2.05 / 2 = 1.025 mol

6) Atomic weight of M:

65.13 g / 1.025 mol = 63.54 g/mol
Copper

Example #9: An oxide of an element with a valance of 6 contains 48% oxygen. What is the atomic weight of this element? Identify the element.

Solution:

An element, M, with a valence of +6 will form an oxide with the formula MO₃.
Let us assume 100 g of the compound is present. Therefore, 48 g is oxygen and 52 g is M.
The moles of oxygen ---> 48 g / 16 g/mol = 3 mol
From the 1:3 molar ratio of M to O, we have this:
1 is to 3 as x is to 3 moles
x = 1 mole (This is how many moles of M are in the 100 g of MO₃.)
52 g / 1 mol = 52 g/mol <--- the atomic weight of M
The element is chromium.

Example #10: An oxybromide compound, KBrO_x, where x is unknown, is analyzed and found to contain 59.19% Br. What is the value of x?

Solution:

1) The three elements exist in the compound in a fixed ratio:

K : Br : O = 1 : 1 : x

2) Use the atomic weights to write an expression for the percent bromine:

59.19 = [(1) (79.9)] / { [(1) (39.1)] + [(1) (79.9)] + [(x) (16.0)] }
59.19 = 79.9 / [119 + 16x]

3) We know x must be an integer, so let us try x = 1:

59.19 = 79.9 / [119 + 16]
79.9 / [119 + 16] = 59.185

4) The formula for KBrO_x is:

KBrO

Problems #1 - 10

Return to Mole Table of Contents

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Determine identity of an element from a binary formula and a percent compositionTen Examples

Determine identity of an element from a binary formula and a percent composition
Ten Examples