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

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


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 / 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.


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 − 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


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% − (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.


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) (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

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) (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

Problem #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


Problem #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

leading to:

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


Problem #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.


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