** Â Â Empirical Formula**

*The simplest formula or the empirical formula provides the lowest whole number ratio of atoms existent in a compound. The relative number of atoms of every element in the compound is provided by this formula.*

**Steps for Determining an Empirical Formula**

**Letâ€™s begin with whatâ€™s given in the problem, i.e., the number of grams of each element.****Weâ€™ll assume that the total mass is 100 grams if percentages are given, so that**

** Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ** **Â Â Each element’s mass = the percentage given.**

**By making use of the molar mass from the periodic table, change the mass of every element to moles.****Divide every mole value by the lowest number of moles computed.****Round up to the closest whole number. Â This is denoted by subscripts in the empirical formula and is the mole ratio of the elements.**

**Multiply each answer by the same factor to get the lowest whole number multiple, if the number is too far to round off (x.1 ~ x.9).**

**e.g. Â Multiply each solution in the problem by 4 to get 5, if one solution is 1.25.**

**e.g. Â Multiply each solution in the problem by 2 to get 3, if one solution is 1.5.**

**The molecular formula can be calculated for a compound if the molar mass of the compound is given when the empirical formula is found. Â To find the ratio between the molecular formula and the empirical formula. Basically, the mass of the empirical formula can be computed by dividing the molar mass of the compound by it. Â Multiply every atom (subscripts) by this ratio to compute the molecular formula.**

**Solved Examples**

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*Problem 1***: A compound contains 88.79% oxygen (O) and 11.19% hydrogen (H). Compute the empirical formula of the compound.**

*Solution:*

**Assume 100.0g of substance. We see that the percent of each element matches the grams of each element**

**Â Â Â Â Â Â Â Â Â Â Â Â Â Â 11.19g H**

**Â Â Â Â Â Â Â Â Â Â Â Â Â Â 88.79g O**

**Convert grams of each element to moles**

**Â Â Â Â Â H: (11.19g) [frac1molHatoms1.008gH] = 11.10 mol H atoms**

**Â Â Â Â Â O: (88.79g) [frac1molOatoms16.00gO] = 5.549 mol O atoms**

**The formula could be articulated as H****11.10****O****5.549****. However, itâ€™s usual to use the smallest whole number ratio of atoms.**

**by dividing the lowest number alter the numbers to whole numbers.**

**Â Â Â Â Â Â Â H = frac11.10mol5.549molfrac11.10mol5.549mol = 2.000**

**Â Â Â Â Â Â Â O = frac5.549mol5.549molfrac5.549mol5.549mol = 1.000**

**Â Â Â Â Â Â The simplest ratio of H to O is 2:1**

**Â Â Â Â Â Â Â Empirical formula = H2O**

**Problem 2: A sulfide of iron was formed by combining 1.926g of sulfur(S) with 2.233g of iron (Fe). What is the compound’s empirical formula?**

**Solution:**

**As the mass of each element is known, we use them directly****Convert grams of each element to moles**

**Fe: (2.233g Fe)(frac1molFeatoms55.85gFe) = 0.03998 mol Fe atoms**

**S: (1.926g S)(frac1molSatoms32.07gS) = 0.06006 mol S atoms**

**By dividing by the smallest number, change the numbers to whole numbers.**

**Fe = frac0.03998mol0.03998mol = 1.000**

**S = frac0.06006mol0.03998mol = 1.502**

**As we still have not reached a ratio that gives whole numbers in the formula we multiply by a number that will give us whole numbers**

**Fe: (1.000)2 = 2.000**

**S: (1.502)2 = 3.004**

**Empirical formula = Fe****2****S****3**