Change moles to grams:
0.62 mol NaOH |
40.01 g NaOH |
1 mol NaOH |
Change grams to moles:
2.5 g NaOH |
1 mol NaOH |
40.01 g NaOH |
Change moles to atoms/molecules:
0.62 mol NaOH |
6.02 x 1023 atoms/mlc NaOH |
1 mol NaOH |
Change atoms/molecules to moles:
2.35 X 1022 atoms/mlc NaOH |
1 mol NaOH |
6.02 X 1023 atoms/mlc NaOH |
Change from atoms/molecules to grams:
2.35 X 1022 atoms/mlc NaOH |
1 mol NaOH |
40.01 g NaOH |
6.02 X 1023 atoms/mlc NaOH |
1 mol NaOH |
Change from grams to atoms/molecules:
2.35 g NaOH |
1 mol NaOH |
6.02 x 1023 atoms/mlc NaOH |
40.01 g NaOH |
1 mol NaOH |
Ba = 1 x 137.33 = 137.33 O = 2 x 16.0 = 48.0 H = 2 x 1.01 = 2.02 _________ 187.35
137.33 ÷ 187.35 = 0.733 x 100 = 73.3% Ba 48.0 ÷ 187.35 = 0.256 x 100 = 25.6% O 2.02 ÷ 187.35 = 0.011 x 100 = 1.1% H
Change % to grams Change grams to moles Divide by the lowest Multiply to get whole #'s 63.1% Mn = 63.1 g Mn x 1 mol Mn = 1.15 ÷ 1.15 = 1 (not necessary in this example) 54.94 g Mn 36.9% S = 36.9 g S x 1 mol S = 1.15 ÷ 1.15 = 1 (not necessary in this example) 32.07 g S Empirical Formula = MnSUse the same procedures for the molecular formula and add the last steps.
Change % to grams Change grams to moles Divide by the lowest Multiply to get whole #'s 92.25% C = 92.25 g C x 1 mol C = 7.68 ÷ 7.67 = 1 (not necessary in this example) 12.01 g C 7.75% H = 7.75 g H x 1 mol H = 7.67 ÷ 7.67 = 1 (not necessary in this example) 1.01 g H Empirical Formula = CH molecular mass = 78 [EFamu]x = molecular mass [13.02]x = 78 (solve for x) 13.02x = 78 x = 6 plug in x and multiply through the empirical formula subscripts [CH]6 = C6H6 = molecular formula