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Calculate empirical formula when given percent composition data

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

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

Determine the formula of a hydrate

**Problem #1:** A sample of copper metal weighing 2.50 g is heated to form an oxide of copper. The final mass of the oxide is 3.13 g. Determine the empirical formula of the oxide.

**Solution:**

1) Determine mass:

Cu ⇒ 2.50 g

O ⇒ 3.13 g - 2.50 g = 0.63 g

2) Determine moles:

Cu ⇒ 2.50 g / 63.546 g/mol = 0.03934 mol

O ⇒ 0.63 g / 16.00 g/mol = 0.039375 molThis is a 1:1 molar ratio between Cu and O.

3) Write the empirical formula:

CuO

**Problem #2:** On analysis, a compound with molar mass 60 g/mol was found to contain 12.0 g of carbon, 2.0 g of hydrogen and 16.0 g of oxygen. What is the molecular formula of the compound? (no calculator allowed!)

**Solution:**

12.0 g carbon is about 1 mole of carbon; 2.0 g of H is about 2 moles and 16.0 g O is about one mole. So the empirical formula is CH_{2}OThe molecular weight of this molecule is 30 g/mole, so you divide 60/30 to find how many times you must multiple your empirical formula. The answer is 2 times the above empirical formula, so the molecular formula is C

_{2}H_{4}O_{2}

**Problem #3:** A compound made of two elements, iridium (Ir) and oxygen (O), was produced in a lab by heating iridium while exposed to air. The following data was collected:

Mass of crucible: 38.26 g

Mass of crucible and iridium: 39.52 g

Mass of crucible and iridium oxide: 39.73 g

**Solution:**

1) Get mass of each element:

Ir ⇒ 39.52 - 38.26 = 1.26 g

O ⇒ 39.73 - 39.52 = 0.21 g

2) Get moles of each element:

Ir ⇒ 1.26 g / 192.217 g/mol = 0.0065551 mol

O ⇒ 0.21 g / 16.00 g/mol = 0.013125 mol

3) Look for smallest whole-number ratio:

Ir ⇒ 0.0065551 / 0.0065551 = 1

O ⇒ 0.013125 / 0.0065551 = 2Empirical formula = IrO

_{2}

By the way, this problem ignores any errors that might be produced by the Ir reacting with the nitrogen in the air.

**Problem #4:** A compound contains 16.7 g of Iridium and 10.3 g of Selenium, what is its empirical formula?

**Solution:**

1) Moles:

Ir ⇒ 16.7 / 192.217 = 0.086881

Se ⇒ 10.3 / 78.96 = 0.130446

2) Seek lowest whole-number ratio:

Ir ⇒ 0.086881 / 0.086881 = 1

Se ⇒ 0.130446 / 0.086881 = 1.5

3) Multiply 1 : 1.5 ratio by two:

2 : 3Ir

_{2}Se_{3}

**Problem #5:** A 1.000 g sample of red phosphorus powder was burned in air and reacted with oxygen gas to give 2.291 g of a phosphorus oxide. Calculate the empirical formula and molecular formula of the phosphorus oxide given the molar mass is approximately 284 g/mol.

**Solution:**

1) Calculate moles of P and O:

P ⇒ 1.000 g / 30.97 g/mol = 0.032289 mol

O ⇒ 1.291 g / 16.00 g/mol = 0.0806875 mol

2) Determine lowest whole-number ratio:

P ⇒ 0.032289 mol / 0.032289 mol = 1

O ⇒ 0.0806875 mol / 0.032289 mol = 2.50

3) Determine empirical formula:

P_{2}O_{5}

3) Determine molecular formula:

The "empirical formula weight" = 141.943284 / 142 = 2

P

_{4}O_{10}

**Problem #6:** A sample of magnetite contained 50.4 g of iron and 19.2 g of oxygen. Calculate the empirical formula.

**Solution:**

1) Convert grams to moles:

Fe ⇒ 50.4 g / 55.845 g/mol = 0.9025 mol

O ⇒ 19.2 g / 16.0 g/mol = 1.2 mol

2) Seek smallest whole-number ratio:

Fe ⇒ 0.9025 / 0.9025 = 1

O ⇒ 1.2 / 0.9025 = 1.33

3) A ratio that involves a third (like the 1.33 just above or something like 2.67) can be thought of this way:

Fe ⇒ 3/3

O ⇒ 4/3

4) Multiply by three to get to whole numbers:

Fe ⇒ 3

O ⇒ 4The empirical formula is Fe

_{3}O_{4}

**Problem #7:** If one molecule with the empirical formula C_{3}H_{7} has a mass of 1.428 x 10^{-22} g, determine the molecular formula of the compound.

**Solution:**

1) Determine mass of one mole:

1.428 x 10^{-22}g/molecule times 6.022 x 10^{23}molecules/mol = 86.0 g/mol

2) Determine "empirical formula weight:"

C_{3}H_{7}= 43.1

3) Determine molecular formula:

86.0 / 43.1 = 2C

_{6}H_{14}

**Problem #8:** What is the empirical formula and molecular formula for lactic acid if the percent composition is 40.00% C, 6.71% H, 53.29% O, and the approximate molar mass is 90 g/mol?

**Solution:**

1) Assume 100 g of the compound is present. This turns percents into mass.

2) Calculate moles:

C ---> 40.00 g / 12.011 g/mol = 3.33

H ---> 6.71 g / 1.008 g/mol = 6.66

O ---> 53.29 g / 16.00 g/mol = 3.33

3) Divide by lowest value:

C ---> 3.33 / 3.33 = 1

H ---> 6.66 / 3.33 = 2

O ---> 3.33 / 3.33 = 1Empirical formula = CH

_{2}O

4) Empirical formula weight

12 + 2 + 16 = 30

5) What is the molecular formula?

90/30 = 3 (there are three "units" of the empirical formula in the molecular formula)C

_{3}H_{6}O_{3}

**Problem #9:** A 0.338 g sample of antimony was completely reacted with 0.295 g of chlorine gas to form antimony chloride. Determine the empirical formula of the antimony chloride.

**Solution:**

Sb ---> 0.338 g / 121.760 g/mol = 0.002776 mol

Cl ---> 0.295 g / 35.453 g/mol = 0.008321 mol0.008321 mol / 0.002776 mol = 3

SbCl

_{3}Note the use of the atomic weight of chlorine, not the molecular weight of Cl

_{2}. We are interested in determining how many Cl atoms bond per one Sb atom, not how many Cl_{2}molecules bond with one Sb. If you use 70.906 g/mol, you get 1.5 chlorine to one Sb (and forgetting that it's 1.5 Cl_{2}molecules) you conclude the formula is Sb_{2}Cl_{3}.

**Problem #10:** When the element antimony, Sb, is heated with excess sulfur, a reaction occurs to give a compound containing only antimony and sulfur. On further heating, excess sulfur is burnt off forming gaseous sulfur dioxide, SO_{2}, and the substance left is pure compound of antimony and sulfur. In one experiment, 2.435 g of antimony was used, and the mass of the pure compound of antimony and sulfur was found to be 3.397 g. What is the empirical formula of the antimony-sulfur compound?

**Solution:**

1) Calculate mass of sulfur reacted:

3.397 - 2.435 = 0.962 g

2) Calculate moles:

Sb ---> 2.435 g / 121.76 g/mol = 0.0200 mol

S ---> 0.962 g / 32.065 g/mol = 0.0300 mol

3) Divide through by smallest:

0.02 / 0.02 = 1

0.03 / 0.02 = 1.5

4) Multiply by 2 to get whole number ratio

Sb_{2}S_{3}

Comment: you can see the 0.2 to 0.3 ratio is the same as a 2 to 3 ratio and arrive at the empirical formula that way. The "divide by smallest" is the classic way to solve problems of this type.

**Problem #11:** A 4.628-g sample of an oxide of iron was found to contain 3.348 g of iron and 1.280 g of oxygen. What is simplest formula for this compound?

**Solution:**

1) Determine moles of each element:

iron: 3.348 g / 55.845 g/mol = 0.05995 mol = 0.06 mol

oxygen: 1.280 g / 16.00 g/mol = 0.08 mol

2) We want the lowest whole number ratio between 0.06 and 0.08:

0.06 / 0.06 = 1 <--- think of it as 3/3

0.08 / 0.06 = 1.3333 <--- think of it as 4/3

3) Multiply by 3:

3/3 x 3 = 3

4/3 x 3 = 4Fe

_{3}O_{4}

Another way to approach this is to see that 0.06 to 0.08 is a 6 to 8 ratio and then reduce it to 3 to 4.

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Calculate empirical formula when given percent composition data

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

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