Simplest ratio formula Ex. CH (Molecular Formula) can be simplified to CH (Empirical Formula) Applies primarly to covalent (molecular) compounds Many compounds have molecular formulas that are the same as their empirical formulas (i.e HO, CO)
Calculating Empirical Formula
Steps:
- Make a mass assumption (most of the time it makes sense to use 100 grams)
- FInd the # of moles (Molar Mass) of each element
- Determine the lowest whole number ratio
Ex. What is the empirical formula of a compound which is found to contain 92.3% C and 7.7% H
Assume 100g of the compound 92.3 grams C and 7.7 grams H moles C: n = m/M = 92.3 g / 12.012 g / mol (Molar Mass of the compound) = 7.683982684 mol C moles H: n = m/M = 7.7 g / 1.008 g/mol (Molar Mass of the compound)= 7.6388888 mol H
Round to Significant Digits C_7$$_.$$_6$$_8 H_7$$_.$$_6$$_2 -> CH
this example results in more of an Empirical Formula since 7.68 and 7.62 can be rounded up to 8
If we had the subscript 6.81 and 18.3 we would divide each subscript by 6.81 and multiply by the correct number to get the empirical formula
The rule we use for rounding is if we are 0. something of a whole number we can round it
Ex. Phosphorous reacts with oxygen to produce a compound that consists of 43.7% phosphorous and 56.3% oxygen. What is the empirical formula of this compound? assume 100g
n = m / M = 43.7g / 30.974 g/mol = 1.410860722 n = m/M = 56.3g / 15.999 g/mol = 3.518969936
-> P(subscript[1.410860722]) O(subscript[3.518969936]) -> P O(subscript[2.494200796]) Multiply the subscript for 2 -> P O(subscript[4.988401592]) Because the subscripts are 0. something to a whole number we can round them to 2 and 5 -> P O
The rule we use for rounding is if we are 0. something of a whole number we can round it
Homework (Grade 11 Chemistry) P.270 # 7 - 12 p.273 # 31 - 40** p.275 # 41 - 50** p.277 # 13 - 18 p.278 # 51 - 60** p.279 # 1 - 16