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

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 64.8% 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

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

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

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% − (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.

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, 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) (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 KBrOx is:

KBrO

Example #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) (39.1)] + [(1) (79.9)] + [(x) (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) (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) (2)]

79.9 / [119 + 32] = 52.914

4) The formula for KBrOx is:

KBrO2

Example #12: A certain metal hydroxide, M(OH)2, contains 32.8% oxygen by mass. What is the identity of the metal M?

Solution:

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 #13: Metal, M, forms an oxide, M2O3, which contains 81.9% of the metal by weight. Find the atomic weight of M.

Solution #1:

1) Assume 100 g of the compound is present.

That means 81.9 g of the compound is M and 18.1 g is O.

2) Determine moles of O:

18.1 g / 15.9994 g/mol = 1.13129 mol

3) The molar ratio between M and O is 2:3. Use this ratio to determine moles of M:

2 is to 3 as x is to 1.13129

x = 0.7541933 mol

4) Determine the atomic weight of M:

81.9 g / 0.7541933 mol = 108.6 g/mol

Solution #2:

1) Molar mass of M2O3 is:

2(M) + 3(O) <--- where M and O are the atomic weights of element M and oxygen

2M + 48.0

2) Set up percent M calculation:

2M / (2M + 48.0) = 0.819

3) Solve:

2M = 0.819(2M + 48.0)

2M = 1.638 M + 39.312

0.362 M = 39.312

M = 108.6 g/mol

Example #14: An ionic compound formed from aluminum and a group VIA element is 18.55 percent Al by mass. What is the formula of the compound?

Solution:

1) The compound will have a generic formula of Al2X3. To solve the problem, we first assume that 100.0 g of the compound is present. 18.55 g of that will be Al. Convert that mass into moles:

18.55 g Al / 26.98 g/mol = 0.6875 mol Al

2) Using the formula of the compound, calculate moles of X:

(0.6875 mol Al) (3 mol X / 2 mol Al) = 1.031 mol X

3) The molar mass of X:

(100.0 − 18.55 g) / 1.031 mol = 79.0.0 g/mol

4) The element in group VIA with a molar mass of 79 g/mol is selenium. So the compound is Al2Se3, named aluminum selenide.

Example #15: An element X forms two oxides whose formulas are XO3 and X2O3. One of the oxides contains 52% X by weight and has a formula weight of 99.98 g/mol. What is the formula of this oxide?

(a) CrO3
(b) Al2O3
(c) Mg2O3
(d) VO3

Solution:

Assume 100 g (or 99.98 g, if you wish) of the oxide is present. This means X accounts for 52 g and O for 48 g.

48 g / 16 g/mol = 3 mol of O present in the 100 g

Therefore, the atomic weight of X will be either 52 (if XO3 works out) or 26 (if X2O3 works out)

CrO3 ---> from the periodic table, the atomic weight of Cr is 52. This is the correct answer. %2 frm the Cr and 48 from the O account for the 00 g in the sample.

Al2O3 ---> The 52 would be divided by 2 to get 26. However, Al is known to weigh 27. This is a wrong answer.

Mg2O3 ---> The 52 would be divided by 2 to get 26. However, Mg is 24. This is a wrong answer.

VO3 ---> the atomic weight of V is 51. Not being 52 makes it a wrong answer.