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EMPIRICAL FORMULAS From Percentages to Formulas CONSTANT COMPOSITION WHAT DOES THAT MEAN? Compounds have constant composition Definite proportions of elements chemically combined within them A chemist could measure this in two different ways: Count numbers of constituent atoms Chemical formula with subscripts would result Example Formula AlCl3 means one atom of aluminum is bonded with three atoms of chlorine Percentages (by mass) of its elements Up until now, you’ve only worked with chemical formulas You believed a chemical formula was just the ratio of number of atoms in a compound Chemical formulas are also the ratio of MOLES of atoms in a compound! In 1 MOLE of CO2, there is 1 MOLE of carbon and 2 MOLES of oxygen TYPES OF CHEMICAL FORMULAS THE EMPIRICAL FORMULA Defined as the lowest whole number ratio of elements in a compound Up until now, you’ve been working with molecular formulas! The actual ratio of elements in a compound C2H4 and C3H6 are both molecular formulas Both have empirical formulas of CH2 Empirical formulas and molecular formulas can be the same! H2 O HOW TO CALCULATE EMPIRICAL FORMULAS GIVEN MOLECULAR FORMULAS Look at molecular formula and find the lowest whole number ratio of atoms Example 1 Molecular: C6H12O6 Empirical: CH2O Example 2 Molecular: CH4N Empirical: CH4N CONSTANT COMPOSITION AND MASS PERCENT It is useful to know how much an element contributes to the total molar mass of a compound So, how do you calculate mass percent? Obtained by comparing the MASS OF EACH ELEMENT present in 1 mole of the compound to the TOTAL MASS of 1 mole of the compound A pure compound should show the same percent mass of each element consistently So given a formula , you should be able to figure out the percent mass of each element BASIC STEPS OF DETERMINING MASS PERCENT FROM FORMULA Write correct formula of compound with subscripts 2. Calculate mass of each element in 1 mole of the compound 3. Get molar mass of compound 4. Find the fraction of the total mass contributed by each element and convert it to a percentage 1. Set up a ratio, or fraction, with molar mass of element on top (numerator) and molar mass of compound on bottom (denominator) 5. Sum the individual mass percent values to make sure they add up to 100%! PRACTICE! Carvone is a substance that occurs in two forms having different arrangements of the atoms but the same molecular formula of C10H14O. One type of carvone gives caraway seeds their characteristic smell, and the other type is responsible for the smell of spearmint oil. Compute the mass percent of each element in carvone. Step 1: Write correct formula of compound with subscripts C10H14O Step 2: Find the masses of each element in 1mole of carvone: 𝑀𝑎𝑠𝑠 𝑜𝑓 𝐶 𝑖𝑛 1 𝑚𝑜𝑙𝑒 = 10 𝑚𝑜𝑙 × 12.01 𝑔 = 120.1 𝑔 1 𝑚𝑜𝑙 𝑀𝑎𝑠𝑠 𝑜𝑓 𝐻 𝑖𝑛 1 𝑚𝑜𝑙𝑒 = 14 𝑚𝑜𝑙 × 1.008 𝑔 = 14.11 𝑔 1 𝑚𝑜𝑙 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑂 𝑖𝑛 1 𝑚𝑜𝑙𝑒 = 1 𝑚𝑜𝑙 × 16.00 𝑔 = 16.00 𝑔 1 𝑚𝑜𝑙 Step 3: Get molar mass of compound 𝑀𝑎𝑠𝑠 𝑜𝑓 1 𝑚𝑜𝑙 𝐶10 𝐻14 𝑂 = 120.1 + 14.11 + 16.00 = 150.2 𝑔 EXAMPLE - CONTINUED Step 4: Find the fraction of the total mass contributed by each element and convert it to a percentage: 𝑀𝑎𝑠𝑠 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑜𝑓 𝐶 = 120.1 𝑔 𝐶 × 100% = 79.96% 150.2 𝑔𝐶10 𝐻14 𝑂 14.11 𝑔 𝐻 𝑀𝑎𝑠𝑠 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑜𝑓 𝐻 = × 100% = 9.394% 150.2 𝑔𝐶10 𝐻14 𝑂 𝑀𝑎𝑠𝑠 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑜𝑓 𝑂 = 16.00 𝑔 𝑂 × 100% = 10.65% 150.2 𝑔𝐶10 𝐻14 𝑂 Step 5: Sum the individual mass percent values to make sure they add up to 100%! STEPS TO USING MASS PERCENTS TO DETERMINE EMPIRICAL FORMULAS 1. Pretend that you have a 100 gram sample of the compound 2. 3. Convert the grams to moles for each element Write the number of each element as a subscript in a chemical formula 4. In other words, change the % to grams Keep each number as a decimal at this point! Divide each subscript by the smallest number This will give the empirical formula 5. Multiply the result by some integer to get rid of any fractions May not be necessary PRACTICE! Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 % N Step 1: Pretend that you have a 100 gram sample of the compound Step 2: Convert the grams to moles for each element EXAMPLE (CONTINUED) Step 3: Write the number of each element as a subscript in a chemical formula C3.22H16.09N3.22 Step 4: If we divide all of these by the smallest subscript, it will give us the empirical formula MORE PRACTICE! Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula? What is the empirical mass? HOW TO CONVERT BETWEEN EMPIRICAL FORMULAS AND MOLECULAR FORMULAS Since the empirical formula is the lowest ratio, the actual molecule would has a bigger mass Molecular formula can always be obtained by multiplying by some whole number To do so, follow the steps below: 1. Divide the actual molecular molar mass (usually given in the problem) by empirical molar mass 2. Gives whole number Multiply empirical formula by the whole number to get the molecular formula Look back at the previous problem. Caffeine has a molar mass of 194 g/mol. What is its molecular formula? PRACTICE! A compound is known to be composed of 71.65 % Cl, 24.27% C, and 4.07% H. Its molecular molar mass is known to be 98.96 g/mol. What is its molecular formula? A REAL-WORLD DETERMINATION OF EMPIRICAL FORMULA REAL-WORLD DETERMINATION OF EMPIRICAL FORMULA Combustion analysis is one of the most common methods for determining empirical formulas A weighed compound is burned in oxygen and its products are analyzed by a gas chromatogram It is particularly useful for analysis of hydrocarbons! Products are CO2 and H2O 16 MORE ON COMBUSTION ANALYSIS Combustion Analysis The technique of finding the mass composition of an unknown sample (X) by examining the products of its combustion X + O2 → CO2 + H2O Example Data 0.250 g of compound X produces 0.686 g CO2 and 0.562 g H2O 17 DETERMINING EMPIRICAL FORMULAS FROM COMBUSTION ANALYSIS X + O2 → CO2 + H2O Step 1: Find the mass of C & H that must have been present in X Multiply masses of products by percent composition (decimal form) of the products C: 0.686 g x (12.01 g/44.01 g) = 0.187 g C H: 0.562 g x [(2 x 1.008 g)/18.02 g]= 0.063 g H 18 COMBUSTION ANALYSIS X + O2 → CO2 + H2O Step 2: Add masses of products and compare to mass of original compound 0.187 g C + 0.063 g H = 0.250 g total Based on original compound mass of 2.90 g, compound X must contain only C and H!! Step 3: Find the number of moles of C and H C: 0.187 g x mole/12.01 g = 0.0156 moles C H: 0.063 g x mole/1.008 g = 0.063 moles H 19 COMBUSTION ANALYSIS X + O2 → CO2 + H2O Step 4: Write the formula using the mole numbers as subscripts C.0156H.063 Step 5: Divide by smallest number of moles C: 0.0156/0.0156 = 1 H: 0.063/0.0156 = 4 If these numbers are fractions, multiply each by the same whole number 20 COMBUSTION ANALYSIS X + O2 → CO2 + H2O Step 6: Rewrite formula with new mole whole numbers You now have the empirical formula! CH4 21