Molar Conversion
1 mol is equal to this number (name the name and number).
Avogadro's number --> 6.022 x 10^23
Define molar mass.
mass in grams of one mole of any pure substance
After dividing the 2 masses (elements/total compound), you must multiply by this number always to get the percent composition.
100
Explain the steps to find empirical formula in your own words.
Convert % into g (# in % is the same in grams)
Convert g into mol and divide all of the elements in mol by the lowest amount of mol
After dividing, if there is still a non-whole number, multiply all the numbers by the smallest number to get the simplest whole-number ratio.
This is the difference between molecular formula and empirical formula.
-empirical: bare bones, simpliest formula
-molecular: usu. more complex, actual formula
This is included in a representative particle.
particle, molecule, atom
Find the molar mass of O2 (2 sig figs after the decimal)
2 mol O (16.00 g/1mol) = 32.00 g O
This is the formula for percent composition.
percent by mass = (mass of element in 1 mol of compound/molar mass of compound) x 100
This is the empirical formula for a compound that has 48.0% C, 4.0% H and 48.0% O
C4H4O3
This is the first step in calculating the molecular formula if empirical formula and molar mass of molecular formula is given.
Find the molar mass of the empirical formula and divide the molar masses.
This is the number of moles in 98.5 grams of gold. (2 sig figs)
98.5 g Au (1 mol/196.97 g) = 0.50 mol Au
Find the molar mass of Mg(OH)2 (2 sig figs after the decimal)
1 mol Mg (24.31 g/1mol) = 24.31 g
2 mol O (16.00 g/1 mol) = 32.00 g
2 mol H (2.02 g/1 mol) = 4.04 g
MM = 24.31 g + 32.00 g + 4.04 g = 60.35 g
This is the percent composition of hydrogen and oxygen in water (4 sig figs overall)
Water = H2O
2 mol H (1.01 g/1 mol) = 2.02 g
1 mol O (16.00 g/1 mol) = 16.00 g
MM = 2.02 + 16.00 = 18.02 g/mol
% H = 2.02/18.02 x 100 = 11.10% H
% O = 16.00/18.02 x 100 = 88.90% O
This is the empirical formula for a compound that has 36.0% C, 4.0% H, 28.0% N and 32.0% O
C3H4N2O2
This is the molecular formula of a compound if its empirical formula is C2OH4 and it has a molar mass of 88 g/mol
MM of MF/MM of EF
= 88/(2(12.01) + 16.00 + 4(1.01))
= 88 / 44.06
= 2 x C2OH4
MF =C4O2H8
This is the number of atoms in 4.189 moles of NaCl. (2 sig figs)
4.189 mol NaCl (6.022 x 1023 / 1 mol) = 2.522 x 1025 atom NaCl
Find the number of atoms of N and H in 3.50 g NH3.
1 mol N (14.00 g / 1 mol) = 14.01 g
3 mol H (1.01 g / 1 mol) = 3.03 g
MM = 17.04 g/mol
3.50 g NH3 (1 mol NH3/17.04 g)(6.022 x 1023 fu/1 mol) = 1.24 x 1023 fu
1.24 x 1023 fu (1 atom N/1 fu NH3) = 1.24 x 1023 atom N
1.24 x 1023 fu (3 atom =H/1 fu NH3) = 3.72 x 1023 atom H
This is the percent composition of (NH4)2S (4 sig figs overall)
2 mol N (14.01 g/1 mol) = 28.02 g
8 mol H (1.01 g/1 mol) = 8.08 g
1 mol S (32.07 g/1 mol) = 32.07 g
MM = 28.02 g + 8.08 g + 32.07 g = 68.17 g
% N = (28.02/68.17) x 100 = 41.20% N
% H = (8.08/68.17) x 100 = 11.80% H
% S = (32.07/68.17) x 100 = 47.04% S
This is the empirical formula for a compound that has 24.0 g C, 7.0 g H, 38.0 g F and 31.0 g P
C2H7F2P
This is the molecular formula for a compound that has an empirical formula of C4H8N2O and a mass of 200 g/mol
MM of MF/MM of EF
= 200/(4(12.01) + 8(1.01) 2(14.01) + 16.00 )
= 200 / 100.14
= 2
MF = 2 x EF
= 2 x C4H8N2O
MF = C8H16N4O2
This is how many atoms are in 100g of potassium hydroxide.
100g (1 mol/56.11 g)(6.022 x 1023 atoms/1 mol) = 1.07 x 1024 atoms KOH
Find the number of atoms of C and O in 2.50 g carbonate.
1 mol C (12.01 g / 1 mol) = 12.01 g
3 mol O (16.00 g / 1 mol) = 48.00 g
MM = 60.01 g/mol
2.50 g CO3 (1 mol CO3/60.01 g)(6.022 x 1023 fu/1 mol) = 2.51 x 1022 fu
2.51 x 1022 fu (1 atom C/1 fu CO3) = 2.51 x 1022 atom C
2.51 x 1022 fu (3 atom O/1 fu CO3) = 7.53 x 1022 atom O
This is the percent composition of NaC2H3O2 (4 sig figs overall)
1 mol Na (22.99 g/1mol) = 22.99 g
2 mol C (12.01 g/1mol) = 24.02 g
3 mol H (1.01 g/1 mol) = 3.03 g
2 mol O (16.00 g/1 mol) = 32.00 g
MM = 22.99 g + 24.02 g + 3.03 g + 32.00 g = 82.04 g/mol
% Na = (22.99/82.04) x 100 = 28.02%
% C = (24.02/82.04) x 100 = 29.28%
% H = (3.03/82.04) x 100 = 3.69%
% O = (32.00/82.04) x 100 = 39.01%
This is the empirical formula of a compound that has 48.0 % C, 8.0 % H, 28.0 % N and 16.0 % O
C4H8N2O
A sample contains 1.388 g of C, 0.342 g H and 1.850 g O, and it’s molecular mass is 62 g. What is the empirical and molecular formula of this substance?
1.388g C (1 mol/12.01 g) = 0.116/0.116=1
0.342g H (1mol/1.01 g) = 0.339/0.116=2.92≈3
1.850 g O x (1 mol O/16.00 g) = 0.116/0.116 = 1
EF = CH3O
MM of MF/MM of EF
= 62/(2(12.01) + 6(1.01) + 2(16.00)) =62/31.04 =2
MF = 2 x EF
=2 x CH3O
MF=C2H6O2