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

Example #1: 100.0 g of XF3 contains 49.2 g of fluorine. What element is X?

Solution #1:

49.2 g / 19 g/mol = 2.589 mol of F

3 is to 2.589 as 1 is to x

x = 0.863 mol of X present

50.8 g / 0.863 mol = 58.8 g/mol

cobalt

Solution #2:

49.2% by weight of your compound is fluorine. In the molecule of your unknown compound you have three atoms of fluorine, this means that in one mole of this compound you have 57 g of fluorine (from 19.0 x 3) that represent 49.2% by weight. One mole of unknown compound weighs (57/0.492) = 115.85 g. Knowing that 57 g are made by fluorine, the unknown atom will count for 58.85 g in each mole. This means that its atomic weight is 58.85 g/mol, the atomic weight of cobalt.

Example #2: A 1.443 g sample of an unknown metal is reacted with excess oxygen to yield 1.683 grams of an oxide known to have the formula M2O3. Calculate the atomic weight of the element M and identify the metal.

Solution:

1) Determine grams, then moles of oxygen:

1.683 − 1.443 = 0.240 g

0.240 g / 16.00 g/mol = 0.015 mol

2) Determine moles of M:

x is to 2 as 0.015 mol is to 3

x = 0.010 mol of M

3) Determine atomic weight of M:

1.443 / 0.010 mol = 144.3 g/mol

Neodymium (element #60)

Example #3: When the element A is burned in an excess of oxygen, the oxide A2O3(s) is formed. 0.5386 g of element A is treated with oxygen and 0.711 g of A2O3 are formed. Identify element A.

Solution:

1) mass oxygen:

0.711 g − 0.5386 g = 0.1724 g

2) mole oxygen:

0.1724 g / 16.00 g/mol = 0.010775 mol

3) moles A:

0.010775 mol is to 3 as x is to 2

x = 0.007183 mol

4) atomic weight (and identity) of A:

0.5386g / 0.007183 mol = 75.0 g/mol

Arsenic

Example #4: A 2.89 g sample of osmium oxide, OsxOy, contains 2.16 g of osmium. What are the values of x and y?

Solution:

1) Determine mass:

Os ⇒ 2.16 g
O ⇒ 2.89 − 2.16 = 0.73 g

2) Divide each by atomic mass:

Os ⇒ 2.16 g / 190.23 g/mol = 0.01135 mol

O ⇒ 0.73 g / 16.00 g/mol = 0.0456 mol

3) Divide by smaller:

Os ⇒ 0.01135 / 0.01135 = 1
O ⇒ 0.0456 / 0.01135 = 4

x = 1
y = 4

Although not asked for, the formula is OsO4

Example #5: 7.8 g of an element X reacts with oxygen to form 9.4 g of an oxide X2O. What is the relative atomic mass of X? What is the element X?

Solution:

1) Mass of O in the compound:

9.4 − 7.8 = 1.6 g

2) Moles of O in the compound:

1.6 g / 16 g/mol = 0.1 mol

3) Determine moles of X in the compound:

From X2O, the molar ratio of X to O is 2 to 1
Our sample contains 0.2 mol of X

4) Determine atomic weight of X:

7.8 g / 0.2 mol = 39 g/mol

X is potassium.

Example #6: A 64.8 g sample of the compound X2O5 contains 48.0 g of oxygen atoms. What is the atomic weight of element X? What element is element X?

Solution:

1) Moles of O in the compound:

48.0 g / 16.0 g/mol = 4.00 mol

2) Moles of X in the compound:

x is to 4 as 2 is to 5

x = 1.60 mol

3) Mass of X in the compound:

64.8 g − 48.0 g = 16.8 g

4) Atomic weight of X:

16.8 g / 1.60 mol = 10.5 g/mol

The element is boron

Example #7: The chloride of an unknown metal is believed to have the formula MCl3. A 1.603 g sample of the compound is found to contain 0.03606 mol of Cl. Determine the atomic weight of element M and identify it by name.

Solution:

(0.03606 mol) (35.453 g/mol) = 1.27844 g

1.603 g − 1.27844 g = 0.32456 g

The M to Cl molar ratio is 1 to 3. Therefore, moles M in the sample:

0.03606 mol / 3 = 0.01202 mol

atomic mass:

0.32456 g / 0.01202 mol = 27.0 g/mol

Aluminum.

Example #8: The 64.8 g sample of the compound X2O5 contains 48.0 grams of oxygen atoms. What is the molar mass of element X? What is the identity of element X?

Solution:

64.8 − 48 = 16.8 g (the mass of X in the sample)

48.0 g / 16.0 g/mol = 3.00 mol of O in sample.

The molar ratio of X to O is 2 to 5.

2 is to 5 as y is to 3.00

y = 1.2 <--- this is the number of moles of X in the sample.

16.8 g / 1.2 mol = 14.0 g/mol

Nitrogen, baby!!

Example #9: A 47.3 g sample of the compound X3(PO4)2 contains 8.78 g of phosphorus. Identity X

Solution #1:

(8.78 / 47.3) * 100 = 18.56% (percentage of P in compound)

and this has molar mass of 2 x 31 g = 62 g

molar mass of the compound ---> 62 x (100 / 18.56) = 334.1 g / mol

molar mass of just PO4 in compound ---> (94.971 x 2 ) g / mol = 189.9 g /mol

molar mass of just the metal ---> (334.1 − 189.9) / 3 = 48.0 g / mol

titanium

Solution #2:

8.78 g / 30.97 g/mol = 0.2835 mol

0.2835 mol as to 2 as y is to 3

y = 0.42525 mol (this is how many moles of X are present in the compound)

0.2835 mol is to 2 as z is to 8

z = 1.134 mol (this is how much oxygen is present)

16.0 g/mol times 1.134 mole = 18.144 g

47.3 g − (8.78 + 18.144) = 20.376 g (grams of X present in the sample)

20.376 g / 0.42525 mol = 47.9 g/mol

Solution #3:

You have 8.78 g of P with an atomic weight of 30.97 g/mole so you have 0.2835 moles.

From the equation, 1 mole of X3(PO4)2 contains 2 moles P so you have 0.2835/2 moles of the mystery compound.

You also know the mass of the compound is 47.3 g so the MW is 47.3 g / 0.14175 moles = 333.7 g/mole

In one mole of the compound, you have 3 moles X, 2 moles P and 8 moles O. Eight moles of O is 128 g and two moles P is 61.9 g.

The mass of the mystery material assuming you had 1 mole = 333.7 g − 128 g − 61.9 g = 143.8.

Since the formula has three X, the atomic weight of X = 47.9 g/mol.

Example #10: A 30.6-g sample of the compound M2O3 contains 14.4 g of oxygen atoms. What is the molar mass of element M? Identify the element M most probably is.

Solution:

1) Set up the following ratio and proportion:

 30.6 14.4 –––––– = –––– 2x + 48 48

14.4 g / 48 g ---> The moles of O in 30.6 g of M2O3

48 g ---> There are 48 g in three moles of oxygen. The three comes from oxygen's subscript in the formula. (I suppose I should write 48 g/3mol.

2x + 48 ---> The molar mas of M2O3. x is the atomic weight of M, note that is multiplied by two because of the subscripted two in the formula.

2) Cross multiply and divide:

(14.4) (2x + 48) = (48) (30.6)

(14.4) (2x + 48) = 1468.8

2x + 48 = 102

2x = 54

x = 27 g/mol

Aluminum

Example #11: An unknown metal M reacts with oxygen to give the metal oxide MO2. Identify the metal based on the following information.

mass of metal: 0.356 g
mass of metal oxide: 0.452 g

Solution:

1) Mass of the oxygen in MO2:

0.452 − 0.356 = 0.096 g

2) Moles of oxygen in MO2:

0.096 g / 16.00 g/mol = 0.0060 mol

3) The moles of M in the sample of MO2:

the molar ratio of M and O in the formula is 1 to 2

1 is to 2 as x is to 0.0060

x = 0.0030 mol

4) Calculate atomic weight and identify M:

0.356 g / 0.0030 mol = 118.7 g/mol

Looking on a periodic table, we find that 118.7 g/mol is the atomic weight of tin.

Example #12: 16.5 g of element X reacts completely with 9.6 g of oxygen to produce a pure sample of XO2. Find the atomic weight and identity of X.

Solution:

9.6 g / 16.00 g/mol = 0.60 mol

mole ratio of X and O in the formula is 1 to 2. Therefore:

1 is to 2 as x is to 0.60

x = 0.30 mol (this is moles of X in the 26.1 g of XO2)

16.5 g / 0.30 mol = 55 g/mol

Manganese

Example #13: An element reacts with bromine to give the bromide, MBr5. If 2.009 g of the element gives 10.648 g of MBr5, what is the element?

Solution:

1) How much Br is present in the sample?

10.648 − 2.009 = 8.639 g

2) How many moles is this?

8.639 g / 79.904 g/mol = 0.108117 mol

3) From the formula, we know that M and Br are in a 1 to 5 molar ratio. We use a ratio and proportion to get the moles of M.

1 is to 5 as x is to 0.108117

x = 0.0216234 mol

4) We now know a mass of M and how many moles that mass is. To get the molar mass of M, we do this:

2.009 g / 0.0216234 mol = 92.9 g/mol

Look for 92.9 on the periodic table and you will find it is the element niobium.

Example #14: A student places 5.00 g of an unknown metal (X) ribbon in a crucible. The crucible is heated until the unknown metal reacted with oxygen to form a white product with the formula X2O3 . The mass of product is determined to be 7.19 g.

(a) What is the molar mass of the unknown metal (X)?
(b) What element is the unknown metal?

Solution:

1) Mass of oxygen:

7.19 − 5.00 = 2.19 g

2) Moles of oxygen:

2.19 g / 16.00 g/mol = 0.136875 mol

3) Use ratio and proportion to determine moles of X:

2 is to 3 as y is to 0.136875

y = 0.09125 mol

4) Determine molar mass of X:

5.00 g / 0.09125 mol = 54.8 g/mol

Manganese.

Example #15: A 30.6-g sample of the compound M2O3 contains 9.79 g of oxygen atoms. What is the molar mass of element M? Identify the element M most probably is.

Solution:

1) Set up the following ratio and proportion:

 30.6 9.79 –––––– = –––– 2x + 48 48

9.79 g / 48 g ---> The moles of O in 30.6 g of M2O3

48 g ---> There are 48 g in three moles of oxygen. The three comes from oxygen's subscript in the formula. (I suppose I should write 48 g/3mol.

2x + 48 ---> The molar mas of M2O3. x is the atomic weight of M, note that is multiplied by two because of the subscripted two in the formula.

2) Cross multiply and divide:

(9.79) (2x + 48) = (48) (30.6)

(9.79) (2x + 48) = 1468.8

2x + 48 = 150

2x = 102

x = 51 g/mol

Bonus Example: A metal X forms two different chlorides. 12.7 g of chloride A contain 7.10 g and 16.3 g of chloride B contains 10.7 g of chlorine. Determine the formula of the compound.

Solution:

1) Determine moles of chloride present in A and B:

moles Cl in A ---> (7.10 g Cl) (1 mol / 35.453 g) = 0.2003 mol
moles Cl in B ---> (10.7 g Cl) (1 mol / 35.453 g) = 0.3018 mol

2) Determine the smallest whole-number ratio between the two chlorine amounts:

Cl in A = 0.2003 / 0.2003 = 1
Cl in B = 0.3018 / 0.2003 = 1.51 = 1.5

Multiply the 1 to 1.5 ratio by two to obtain the smallest whole-number ratio of Cl in A to Cl in B as 2 to 3.

3) Determine the mass of X in A and B:

mass X in A ---> 12.7 − 7.10 = 5.6 g
mass of X in B ---> 16.3 − 10.7 = 5.6 g

4) In both A and B, there are equal masses, therefore equal number of moles of X. This allows us to determine a partial formula for compounds A and B:

XzCl2
XzCl3

5) Let us speculate about z by considering compound A:

0.2 moles of Cl are present

Assume a 1:2 ratio of X to Cl <--- For compound B, it would be to assume a 1:3 ratio

0.1 mol of X is present

5.6 g / 0.1 mol = 56 g/mol

A reasonable conclusion about X is that it is iron.

6) With z = 1, the two formulas would be:

FeCl2
FeCl3

7) Here is another solution to this problem. It includes discussion about the consequences of assuming a 1:1 ratio between X and Cl. You might also be interested in this solution.

8) In my notes, I also found a solution that utilizes the Law of Multiple Proportions.

 mass of element 1 (compound B) ––––––––––––––––––––––––––– mass of element 2 (compound B) –––––––––––––––––––––––––––––––––– = a small number ratio mass of element 1 (compound A) ––––––––––––––––––––––––––– mass of element 2 (compound A)

9) Substituting:

 10.7 g ––––– 5.6 g 1.5 –––––––––––––– = –––––– 7.1 g 1 ––––– 5.6

10) Here is a statement of the Law of Multiple Proportions:

"When two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers"

11) From the analysis in step 9, we can conclude the following:

For every 1 Cl in compound A, there are 1.5 Cl in compound B. Therefore, based upon Dalton's atomic theory, there are two Cl atoms in compound A for every three Cl atoms in compound B.