### Determine identity of an element from a binary formula and a percent compositionProblems #1 -10 (incomplete)

Problem #1: 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 #2: An oxide of an element has the formula M2O7 and contains 50.45 % oxygen. (a) What is the gram-atomic weight of the element? (Identify the most probable element for M to be.)

Solution to (a):

1) Assume 100 g of the oxide is present. That means 50.45 g of O and 49.55 g of M are present.

2) Determine moles of O:

 1 mol O 50.45 g O x ––––––– =  3.153 mol O 16.00 g

3) Determine moles of M:

 2 mol M x ––––––– = –––––––––– 7 mole O 3.153 mol O

x = 0.90086 mol M

4) Gram-atomic weight of M:

49.55 g / 0.90086 mol = 55.00 g/mol

Solution to (b):

On a periodic table, we find that manganese weighs 54.94 g/mol.

In addition, Mn2O7 is known to exist.

Therefore, we conclude that the unknown element is most probably manganese.

Problem #3: 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

Problem #4: 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.

Problem #5: 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. 52 from the Cr and 48 from the O account for the 100 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.

Problem #6: A metal (M) forms a compound with the formula MCl3. If the compound contains 65.57% Cl by mass, what is the identity of the metal?

Solution:

1) Assume 100 g of the compound is present. This turns our percents into mass. Let us calculate moles:

Cl ---> 65.57 g / 35.543 g/mol = 1.8448 mol

2) From the formula, the molar ratio of M to Cl is 1:3. How many moles of M are present?

1 is to 3 as x is to 1.8448 mol

x = 0.61493 mol

3) We know that there was 34.43 g of M present in our 100 g of MCl3. We can now determine the atomic weight of M:

34.43 g / 0.61493 mol = 56.0 g/mol

This is very close to the atomic weight of iron, which is 55.845 g/mol

Our compound is FeCl3.

Problem #7: A binary compound between an unknown element X and hydrogen contains 92.258% X and 7.742% hydrogen by mass. If the formula of the compound is X4H4, calculate the atomic mass of X and identify what element it is.

Solution:

1) Assume 100 g of X4H4 is present. This means 92.258 g of X and 7.742 g of H are present.

2) Determine moles of H present:

H ---> 7.742 g / 1.008 g/mol = 7.68056 mol

3) There is a 1:1 molar relationship between X and H. Therefore, 7.68056 mol of X is present. Determine the molar mass of X:

92.258 g / 7.68056 mol = 12.01 g/mol

Carbon.

Problem #8: Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.

Solution:

1) Assume 100 grams of the compound is present. That means 69.9 g is Fe and 30.1 g is dioxygen. (The fact that it is dioxygen will be important just below.)

2) Determine moles of iron and dioxygen

iron ---> 69.9 / 55.845 = 1.2517 mol of Fe
dioxygen ---> 30.1/31.9988 = 0.94066 mol of dioxygen

3) However, the empirical formula of a chemical compound is the simplest whole-number ratio of atoms present in a compound. Dioxygen is not O, it is O2. Rewrite the mole amounts:

iron ---> 69.9 / 55.845 = 1.2517 mol of Fe
oxygen ---> 30.1/15.9994 = 1.88132 mol of oxygen atoms

two atoms of O for every dioxygen (O2)

therefore:

(0.94066 mol O2) (2 mol O / 1 mol O2) = 1.88132 mol O

5) Divide through by lowest:

iron ---> 1.2517 1.2517 = 1
oxygen ---> 1.88132 / 1.2517 = 1.5

6) The empirical formula is Fe2O3.

7) This link goes to a discussion of this problem on an "answers" website. Some of the responses miss the point about the use of 'dioxygen' in the question. The writer put it there deliberately to create havoc with the student (as seen in the discussion by the original poster. The idea is to get the student to see that the empirical formula is the smallest whole-number ratio of atoms in the formula, not molecules.

Problem #9: A compound M2O3 contains 68% by mass of element X. Determine what element M is.

Solution:

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

68 g of M is present
32 g of O is present

2) Determine moles of oxygen:

32 g / 16 g/mol = 2.0 mol

3) There is a 2:3 molar ratio between M and O. Therefore:

2 is to 3 as x is to 2.0

x = 1.3333 mol of M

4) Calculate atomic mass of M and identify the element:

68 g / 1.3333 mol = 51 g/mol

Problem #10: The vapor density of a metal chloride is 66.7. Its oxide contains 53% metal. Determine the most probable identity of the metal.

Solution:

1) The vapor density is approx half that of the molar mass. See here.

2) Our MClx weighs approx 133.4 g/mol.

3) Let us guess that the chloride is AlCl3. This means the three Cl weigh 106.5 and the Al weighs 26.9.

4) Let us determine if Al2O3 has 53% of the metal. Go here to determine that Al2O3 does have 53% Al.

5) There is no set of formulas to solve this problem (unless I missed something, which could be the case). A guess must be made as to the identity of the element and calculations like the above be done in order to determine if the guess was the correct one. Aluminum was not my first guess.