Mass Percent, Empirical Formula and Molecular Formula

Mass Percent (percent composition) percent of an element in a given compound Mass percent = Total mass of element present in the compound x 100 Total mass of the entire compound (Molar Mass)

Find the mass percent of each element in ethanol (C2H5OH) First, calculate the molar mass of the compound: 2 C = 2 x 12.01g = 24.02g 6 H = 6 x 1.01g = 6.06g 1 O = 1 x 16.00g = 16.00g 46.08g/mol

Find the mass percent of each element in ethanol (C2H5OH) Then, find the ratio of each element’s mass to the total mass of the compound Mass % of C = 24.02g x 100 = 52.13 % 46.08g Mass % of H = 6.06g x 100 = 13.2 % Mass % of O = 16.00g x 100 = 34.72 %

Find the mass percent of each element in ethanol (C2H5OH) NOTE: The label for mass percent is simply “%”.   Check your work - If you’ve done this correctly, all the percentages should add up to (approximately) 100%! It might be slightly off due to rounding from significant figures. 52.13% + 13.2% + 34.72% = 100.1% (close enough to 100 % !!)

Ex. Find the mass percent of each element in xenon tetrafluoride

Ex. Find the mass percent of each element in copper (II) carbonate (CuCO3)

Molecular Formula gives the actual whole-number ratio of each element present  Ex. The molecular formula for glucose is C6H12O6 A molecule of glucose contains:   6 carbon atoms 12 hydrogen atoms and 6 oxygen atoms

Empirical Formula gives the lowest whole-number ratio of each element present Ex. The empirical formula for glucose is CH2O   A molecule of glucose always has the ratio of 1 carbon atom : 2 hydrogen atoms : 1 oxygen atom

Ex. Determine the empirical formula of the following molecular formulas C6H6 → C12H4Cl4O2 → C6H16N2 → C2H2 → N2O4 → H2O → CH   C6H2Cl2O C3H8N NO2 H2O NOTICE: Which two molecular formulas above have the same empirical formula?  

Calculation of Empirical Formulas The FIRST STEP you must do to calculate empirical formulas is to determine the mass of each element. The information can be given to you in different ways: Directly stated: Ex. The compound is composed of 0.2636 grams of nickel and 0.0718 grams of oxygen. Indirectly stated: The mass of one element and the total mass of the compound is stated. You must subtract to get the mass of the remaining element. Ex. 0.2636 grams of nickel is burned (combined with oxygen). The total mass of the nickel oxide compound is 0.3354 grams. You must subtract 0.2636 g (mass of nickel alone) from 0.3354 g (total mass of nickel and oxygen) to get 0.0718 g of oxygen.

Calculation of Empirical Formulas Given as percentages: Since all the parts together equal 100%, each percentage given can be thought of as grams of that element. (In other words, just change the “%” sign to a “g” for grams.   Ex. The sample was tested and determined to be 78.6% nickel and 21.4% oxygen. Use 78.6 grams of nickel and 21.4 grams of oxygen. Indirectly stated and presented as percentages: Ex. The nickel oxide compound was found to be 78.6% nickel. Only nickel and oxygen are in the compound. If 78.6% nickel, subtract from 100% to realize 21.4% is oxygen. Then, use 78.6 grams of nickel and 21.4 grams of oxygen.

Calculation of Empirical Formulas The SECOND STEP is to convert each mass into moles using molar masses from the periodic table. The THIRD STEP is to calculate a ratio of the moles of one element to moles of the other by dividing all of the mole values by the smallest mole value.

Calculation of Empirical Formulas Sometimes you have to do a FOURTH STEP when step #3 does not give you small, whole number subscripts. IF necessary, multiply all the values from the step #3 by a small whole number (usually between 2 and 6) to make all subscripts into whole numbers, while keeping the relative ratio the same.   For example, if the values are 1 mole aluminum: 1.5 mole oxygen, you obviously cannot have a compound with a formula of Al1O1.5 Multiply by a “2” to remove the fraction and get a formula of Al2O3 For example, if the values above are 1 mole lead: 1.33 mole phosphorus, you obviously cannot have a compound with a formula of Pb1P1.33 Multiply by a “3” to remove the fraction and get a formula of Pb3P4

Calculation of Empirical Formulas Helpful Hint: Remember your fractions! 1/2 = .5 1/3 = .33 1/4 = .25 2/3 = .66 3/4 = .75 So then we multiply by the denominator of the fraction! X 2 X 3 X 4

Calculation of Empirical Formulas FIFTH STEP - Once you have the subscripts, make sure to write the actual empirical formula. Students often forget this and lose points since the point of the problem is to determine an empirical formula, not just determine subscripts.   HELPFUL HINT: Often students get done with step 2 and cannot remember what to do next. A helpful structure, as seen in the attached examples, goes through calculating the values of mass  moles  ratio  subscript

  MASS MOLES RATIO SUBSCRIPT Element A Element B Element C, etc.

Example #1 An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen. Calculate the empirical formula for this compound. Find the mass of each element present. In this example, they are directly stated. Convert each mass into moles. 4.151 g Al x 1 mole Al = 0.1539 mol Al 1 26.98 g Al 3.692g O x 1 mole O = 0.2308 mol O 1 16.00 g O   MASS MOLES RATIO SUBSCRIPT aluminum 4.151 g oxygen 3.692 g

Example #1 3. The third step is to calculate a ratio of the moles of one element to moles of the other by dividing all of the mole values by the smallest mole value. 0.1539 mole of aluminum is the smallest value. Divide both mole values by this. 0.1539 mole Al = 1.000 0.1539 mole Al 0.2308 mole O = 1.500   MASS MOLES RATIO SUBSCRIPT aluminum 4.151 g 0.1539 oxygen 3.692 g 0.2308

Example #1 4. Look at the values from step 3. You must multiply by a “2” to remove the fraction.   MASS MOLES RATIO SUBSCRIPT aluminum 4.151 g 0.1539 1.000 X 2 = 2 oxygen 3.692 g 0.2308 1.500 X 2 = 3

Example #1 This makes the empirical formula as follows NOTE: remember to obey nomenclature rules and write the metal first and the non-metal second ANSWER IS:   Al2O3

Example #2 A student weighs 0.2636 grams of nickel and places it in a crucible. The nickel is then burned (combined with oxygen from the air). The resulting nickel oxide compound weighs 0.3354 grams. What is the empirical formula of the nickel oxide compound?

Example #2 1. Find the mass of each element present. In this example, they are indirectly stated: You must subtract 0.2636 g (mass of nickel alone) from 0.3354 g (total mass of nickel and oxygen) to get 0.0718 g of oxygen.   MASS MOLES RATIO SUBSCRIPT nickel 0.2636 g oxygen 0.0718 g

Example #2 2. Convert each mass into moles.    0.2636 g Ni x 1 mole Ni = 0.004491 mol Ni 1 58.70 g Ni    0.0718 g O x 1 mole O = 0.00449 mol O 1 16.00 g O   MASS MOLES RATIO SUBSCRIPT nickel 0.2636 g 0.004491 oxygen 0.0718 g 0.00449

Example #2 3. The third step is to divide all of the mole values by the smallest mole value. 0.00449 mole of oxygen is the smallest value. Divide both mole values by this. 0.004491 mole Ni = 1.00 0.00449 mole O 0.00449 mole O = 1.00   MASS MOLES RATIO SUBSCRIPT nickel 0.2636 g 0.004491 1.00 oxygen 0.0718 g 0.00449

Example #2 4. Look at the values from step 3. These numbers are extremely close to a whole number and can be simply rounded into subscripts of 1 for nickel and 1 for oxygen. No multiplication is needed.   MASS MOLES RATIO SUBSCRIPT nickel 0.2636 g 0.004491 1.00 1 oxygen 0.0718 g 0.00449

Example #2 NiO This makes the empirical formula as follows: As always: subscripts are assumed to be one unless otherwise indicated Remember: to obey nomenclature rules and write the metal first and the non-metal second   NiO

Empirical Formula Practice Problem A compound contains: 58.84% barium (Ba), 13.74% sulfur and the rest is oxygen. What is the empirical formula. SO THE ANSWER IS: BaSO4 Day 2   MASS MOLES RATIO SUBSCRIPT Barium Sulfur Oxygen 58.84 13.74 27.42 / 137.33 = / 32.06 = / 16.00 = 0.4285 0.4286 1.714 / 0.4285 = 1.00 4.00 * 1= * 1 = 1 4

Do Now Caffeine is made up of 49.48% carbon, 5.19% hydrogen, 16.48% oxygen and 28.8% nitrogen. Find the empirical formula. ANSWER IS : C4H5ON2

Molecular Formulas We’ve learned to calculate empirical formulas. This is the smallest whole-number ratio of the atoms in a compound. They are based on tests that determine the relative masses of the various elements in a compound. Now, we will determine the molecular formula, the actual whole-number ratio of the atoms in the compound. To do this, additional test results or information is needed which tells the molar mass of the molecular formula.

Molecular Formulas The ratio of the molecular formula to the empirical formula must be a whole number. You can remember this by using the acronym “men” (sorry ladies!! ). molecular formula = number OR think m = n empirical formula e

Example 1 If you know the molecular formula and the empirical formula, you can calculate the number. C12H4Cl4O2 = number = 2 (the molecular formula is 2x as big C6H2Cl2O as the empirical formula)

Example 2 If you know the empirical formula and the number and you can calculate the molecular formula.      molecular formula = 6 This means the molecular formula is 6X CH as big as the empirical formula. The molecular formula is C6H6

Example Problem Solving Tests have determined the empirical formula of a compound to be CH4 and the molar mass of the molecular formula to be between 78 and 82 grams per mole. Determine the molecular formula of the compound.

Approach: Find enough information to use. The first piece of information given is that the empirical formula is CH4. The second piece of information gives you the molar mass of the molecular formula. These tests are often not very precise. If you are given a range for the molar mass of the molecular formula, do your calculations using the midpoint of the range. In this case, it would be 80. grams per mole.

Solve The Problem: Find enough information to use:    molecular formula molar mass = number OR think m = n empirical formula molar mass e You are given the molecular formula molar mass (m). You are also given the empirical formula. Calculate the molar mass of the empirical formula (e) and then solve for n.

Solve CH4 x 5 = C5H20 Multiply the empirical formula by the number: Molar mass of CH4 (the empirical formula)= 1 C: 1 x 12.01 g = 12.01 g 4 H: 4 x 1.01 g = 4.04 g 16.05 g/mol   Solve for n: molecular formula molar mass = number OR think m = n empirical formula molar mass e 80. g/mol  5 16.05 g    Multiply the empirical formula by the number: CH4 x 5 = C5H20

Example Problem Solving 2 The empirical formula of a compound is P2O5 and its molar mass is between 280-285 grams. What is the molecular formula? P : 2 x 30.97 = 61.94 O : 5 x 16.00 = +80.00 141.94 282.5 = 1.99 Which is basically 2 So the formula has to be multiplied by 2 making it P4O10

Revisit the caffeine problem Caffeine is made up of 49.48% carbon, 5.19% hydrogen, 16.48% oxygen and 28.8% nitrogen. Find the molecular formula if it’s molar mass is 194.22 remembering that the empirical formula was C4H5ON2. Answer : C8H10O2N4

Review  Percent Composition Determine the mass percent of each element present in iron (III) oxide.

Review Empirical Formula A compound contains 65.95% barium and 34.05% chlorine by mass. Calculate the empirical formula.

Review Empirical Formula A 4.01 g sample of mercury is strongly heated in air and the resulting oxide weighs 4.33 g. Calculate the empirical formula.

Review Molecular Formula The empirical formula of a compound is CH2. The molar mass is approximately 84 g. Determine the molecular formula.

Empirical & Molecular Formula A compound consists of 46.91% sodium, 24.51% carbon, and 28.58% nitrogen. The molar mass of the compound is approximately 50. grams. Determine the empirical formula and the molecular formula of the compound.