empirical and molecular/exact formula

textbook chapter 6.2, april 23-24 lesson, 2025

empirical formula

• simplest ratio of elements in a chemical formula

• mot the exact number in a molecule or formula unit

• determined by measuring mass in a chemical reaction and precent composition

steps of finding an empirical formula, percentage of each element given

1. write out formula with place holders

2. write out mass ratio

3. find n by pretending that the mass is 100 g (precent = mass)

4. to get whole numbers, divide each n by lowest term

5. round to whole number if .0 or .99 is achieved

6. if not, multiply to get a whole number (view decimals and fractions card)

determine the empirical formula of a substance that is 40.0% carbon, 7.0% hydrogen, 53.0% oxygen

1. C_xH_yO_z

2. mass ratio, assuming there is 100 g of substance, is 40:7:50

3. n (mol) = m (g) / mm (g/mol):
   n_C = 40.0 g / 12.011 g/mol = 3.33 mol
   n_H = 7.0 g/ 1.00794 g/mol = 6.944 mol
   n_O = 53.0 g/15.9994 g/mol = 3.312 mol
   (keep all decimals, shortened here for space)

4. oxygen has lowest number of moles, divide by oxygen
   C = 3.33 mol/ 3.312 mol = 1.005
   H = 6.944 mol/ 3.312 mol = 2.096
   O = 3.312 mol/ 3.312 mol = 1

5. C = 1, H = 2, O = 1, therefore the empirical formula is CH₂O

important note: is this the real molecular formula? Maybe, it could be any form of the ratio such as CH₂O, C₂H₄O₂, C₃H₆O₃, etc. bringing from higher infinite options to lower infinite options.

determine the empirical formula for the compound that is 81.7% C and 18.3% H

1. C_xH_y

2. 81.7 : 18.3

3. n_C = 81.7 mol/ 12.011 g/mol = 6.802098077
   n_H = 18.3 mol/ 1.00794 g/mol = 18.15584261

4. C = 6.802098077 mol/ 6.802098077 mol = 1
   H = 18.05584261 mol/ 6.802098077 mol = 2.669153312

5. because H does not equal a .0 or .99, we multiply all elements by 3 to get a whole number (.66 = 2/3, multiply by denominator)

6. C = 1 × 3 = 3
   H = 2.669153312 × 3 = 8.007459936 = 8

7. therefore the empirical formula is C₃H₈

decimals and their corresponding fractions

• .66 = 2/3

• .2 = 1/5

• .25 = ¼

• .33 = 1/3

• .5 = ½

• .75 = ¾

• .125 = 1/8

• if not sure, put number over 100 and simplify

molecular/exact formula

• exact ratio in formula

• molecular formula is a multiple of empirical formula

steps of finding molecular formula

method one: empirical to molecular

1. determine empirical formula

2. molecular formula = X(empirical formula), find X by doing X = mm_molecular/mm_empirical this answer MUST be in whole number, if not check empirical formula

3. multiple empirical formula by X to get moleculer

4. therefore statement of final formula

method 2: directly molecular, if mass given

1. determine number of moles of each element using mass given
   if percentage and formula mass given, multiply in decimal form to get amount of each
   if whole mass and part mass, subtract part from whole to get other part

2. therefore statement of final formula

determine the molecular formula of the compound with the empirical formula CH₂O and molecular mass of 180 amu

method 1: since molecular formula is multiple of empirical formula, so molecular mass (180 amu) is also multiple

1. empirical = CH₂O

2. X = 180 g/mol / 30 g/mol = 6

3. CH₂O x 6 = C₆H₁₂O₆

4. therefore the final formula is C₆H₁₂O₆

find the exact formula of the compound that is 21.9% Na, 45.7% C, 1.9% H, 30.5% O, with a formula mass of 210 amu

method 2

1. in one mole of substance, mass is 210 g
   n_Na = (0.219)(210 g)/ 22.98977 g/mol = 2 mol
   n_C = (0.457)(210 g)/ 12.011 g/mol = 8 mol
   n_H = (0.019)(210 g)/ 1.00794 g/mol = 4 mol
   n_O = (0.305)(210 g)/ 15.9994 g/mol = 4 mol

2. therefore the exact formula is Na₂C₈H₄O₄

a 50.0 g sample a hydrate consist of 27.2 g of anhydrous barium hydroxide salt. Determine the exact formula of the hydrate

method 2

givens: M_hydrate = 50.0 g, M_anhydrate = 27.2 g, formula = Ba(OH)₂⋅XH₂O → find X

need to find ratio of # of moles of water to # of moles anhydrate

1. n_H2O / n_anhydrate= x
   n_anhydrate = m_anhydrate/ mm_anhydrate = 27.2 g/ 121.34468 g/mol = 0.1587443509
   n_H2O = m_H2O/ mm_H2O = 22.8 g [got this number doing 50.0 - 27.2] / 18.01528 g/mol = 1.265592219 mol
   then divide by lowest term:
   n_anhydrate = 0.1587443509 mol/ 0.1587443509 mol = 1
   n_H2O = 1.265592319 mol/ 0.1587443509 mol = 8

2. therefore the formula is Ba(OH)₂⋅8H₂O
   

hydrate

in ionic crystal lattice, small spaces between ions are created. Water molecules, which have polarity, are attracted to the oppositely charged ions and fit into the small spaces

anhydrous

ionic compound with no water

what is Ba(OH)₂ in Ba(OH)₂⋅XH₂O

anhydrate

what is H₂O in Ba(OH)₂⋅XH₂O

waters of hydration