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

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Calculate empirical formula when given mass data

Calculate empirical formula when given percent composition data

Determine identity of an element from a binary formula and mass data

Determine the formula of a hydrate


Problem #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


Problem #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 divided by 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 minus 64.130 = 95.94 g/mol

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


Problem #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 minus 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 minus 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


Problem #4: A compound whose empirical formula is XF3 consists of 64.8% F by mass. What is the atomic mass of X?

Solution:

64.8 g / 19.0 g/mol = 3.4105 mol

y is to 1 as 3.4105 is to 3

y = 1.137 mol of X

35.2 g / 1.137 mol = 30.96 g/mol


Problem #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.


Problem #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


Problem #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.


Problem #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

The metal is 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% minus (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 minus 34.02 = 63.54 g/mol

Copper.


Problem #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.


Problem #10: An oxybromide compound, KBrOx, 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 times 79.9) / [(1 times 39.1) + (1 times 79.9) + (x times 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 KBrOx is:

KBrO

Problem #11: An oxybromide compound, KBrOx, where x is unknown, is analyzed and found to contain 52.92% 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:

52.92 = (1 times 79.9) / [(1 times 39.1) + (1 times 79.9) + (x times 16.0)]

52.92 = 79.9 / [119 + 16x)

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

52.92 = 79.9 / [119 + (16 times 1))

79.9 / [119 + 16) = 59.185 (which does not equal 52.92)

4) Let us try x = 2:

52.92 = 79.9 / [119 + (16 times 2))

79.9 / [119 + 32) = 52.914

4) The formula for KBrOx is:

KBrO2

Return to Mole Table of Contents

Calculate empirical formula when given mass data

Calculate empirical formula when given percent composition data

Determine identity of an element from a binary formula and mass data

Determine the formula of a hydrate