Empirical Formula worksheet

1. Empirical formula: H₄₀₂₈Eu₅Lv₁₃Mc₃

1. Turn the grams of H into moles (a count of atoms): divide by its atomic weight, 1.008 g/mol

   $\frac{42\text{\hspace{0.17em}g}}{1.008\text{\hspace{0.17em}g/mol}}$$=41.666667\text{\hspace{0.17em}mol}$

2. Turn the grams of Eu into moles (a count of atoms): divide by its atomic weight, 151.96 g/mol

   $\frac{7.8\text{\hspace{0.17em}g}}{151.96\text{\hspace{0.17em}g/mol}}$$=0.051329\text{\hspace{0.17em}mol}$

3. Turn the grams of Lv into moles (a count of atoms): divide by its atomic weight, 293 g/mol

   $\frac{40\text{\hspace{0.17em}g}}{293\text{\hspace{0.17em}g/mol}}$$=0.136519\text{\hspace{0.17em}mol}$

4. Turn the grams of Mc into moles (a count of atoms): divide by its atomic weight, 290 g/mol

   $\frac{9\text{\hspace{0.17em}g}}{290\text{\hspace{0.17em}g/mol}}$$=0.031034\text{\hspace{0.17em}mol}$

5. Divide every mole amount by the smallest (0.031034 mol). That makes the scarcest element 1 and shows the ratio

   $\text{H 1,342.593 : Eu 1.654 : Lv 4.399 : Mc 1}$

6. Some ratios aren't whole numbers, so multiply them all by 3. Atoms only come in whole numbers

   $\text{H 4,027.778 : Eu 4.962 : Lv 13.197 : Mc 3}$

7. No small multiplier gives near-whole numbers, so this is the closest fit. Check the data

   $\text{H 4028 : Eu 5 : Lv 13 : Mc 3 → H₄₀₂₈Eu₅Lv₁₃Mc₃}$

2. Empirical formula: Br₆₂Mn₁₆Pt₂₄Gd₇

1. Turn the grams of Br into moles (a count of atoms): divide by its atomic weight, 79.904 g/mol

   $\frac{45\text{\hspace{0.17em}g}}{79.904\text{\hspace{0.17em}g/mol}}$$=0.563176\text{\hspace{0.17em}mol}$

2. Turn the grams of Mn into moles (a count of atoms): divide by its atomic weight, 54.938 g/mol

   $\frac{8\text{\hspace{0.17em}g}}{54.938\text{\hspace{0.17em}g/mol}}$$=0.145619\text{\hspace{0.17em}mol}$

3. Turn the grams of Pt into moles (a count of atoms): divide by its atomic weight, 195.08 g/mol

   $\frac{42\text{\hspace{0.17em}g}}{195.08\text{\hspace{0.17em}g/mol}}$$=0.215296\text{\hspace{0.17em}mol}$

4. Turn the grams of Gd into moles (a count of atoms): divide by its atomic weight, 157.25 g/mol

   $\frac{10\text{\hspace{0.17em}g}}{157.25\text{\hspace{0.17em}g/mol}}$$=0.063593\text{\hspace{0.17em}mol}$

5. Divide every mole amount by the smallest (0.063593 mol). That makes the scarcest element 1 and shows the ratio

   $\text{Br 8.856 : Mn 2.29 : Pt 3.386 : Gd 1}$

6. Some ratios aren't whole numbers, so multiply them all by 7. Atoms only come in whole numbers

   $\text{Br 61.992 : Mn 16.029 : Pt 23.699 : Gd 7}$

7. No small multiplier gives near-whole numbers, so this is the closest fit. Check the data

   $\text{Br 62 : Mn 16 : Pt 24 : Gd 7 → Br₆₂Mn₁₆Pt₂₄Gd₇}$

3. Empirical formula: Ts₁₂At₃Ag₂₉Br₁₀

1. Take a 100 g sample, so each percentage becomes that many grams

   $\text{Ts: 55 g, At: 10 g, Ag: 50 g, Br: 13 g}$

2. Turn the grams of Ts into moles (a count of atoms): divide by its atomic weight, 294 g/mol

   $\frac{55\text{\hspace{0.17em}g}}{294\text{\hspace{0.17em}g/mol}}$$=0.187075\text{\hspace{0.17em}mol}$

3. Turn the grams of At into moles (a count of atoms): divide by its atomic weight, 210 g/mol

   $\frac{10\text{\hspace{0.17em}g}}{210\text{\hspace{0.17em}g/mol}}$$=0.047619\text{\hspace{0.17em}mol}$

4. Turn the grams of Ag into moles (a count of atoms): divide by its atomic weight, 107.87 g/mol

   $\frac{50\text{\hspace{0.17em}g}}{107.87\text{\hspace{0.17em}g/mol}}$$=0.463521\text{\hspace{0.17em}mol}$

5. Turn the grams of Br into moles (a count of atoms): divide by its atomic weight, 79.904 g/mol

   $\frac{13\text{\hspace{0.17em}g}}{79.904\text{\hspace{0.17em}g/mol}}$$=0.162695\text{\hspace{0.17em}mol}$

6. Divide every mole amount by the smallest (0.047619 mol). That makes the scarcest element 1 and shows the ratio

   $\text{Ts 3.929 : At 1 : Ag 9.734 : Br 3.417}$

7. Some ratios aren't whole numbers, so multiply them all by 3. Atoms only come in whole numbers

   $\text{Ts 11.786 : At 3 : Ag 29.202 : Br 10.25}$

8. No small multiplier gives near-whole numbers, so this is the closest fit. Check the data

   $\text{Ts 12 : At 3 : Ag 29 : Br 10 → Ts₁₂At₃Ag₂₉Br₁₀}$

4. Empirical formula: Pd₇Si₅Ag₉Sr₂

1. Take a 100 g sample, so each percentage becomes that many grams

   $\text{Pd: 43 g, Si: 7.8 g, Ag: 54 g, Sr: 10 g}$

2. Turn the grams of Pd into moles (a count of atoms): divide by its atomic weight, 106.42 g/mol

   $\frac{43\text{\hspace{0.17em}g}}{106.42\text{\hspace{0.17em}g/mol}}$$=0.404059\text{\hspace{0.17em}mol}$

3. Turn the grams of Si into moles (a count of atoms): divide by its atomic weight, 28.085 g/mol

   $\frac{7.8\text{\hspace{0.17em}g}}{28.085\text{\hspace{0.17em}g/mol}}$$=0.277728\text{\hspace{0.17em}mol}$

4. Turn the grams of Ag into moles (a count of atoms): divide by its atomic weight, 107.87 g/mol

   $\frac{54\text{\hspace{0.17em}g}}{107.87\text{\hspace{0.17em}g/mol}}$$=0.500603\text{\hspace{0.17em}mol}$

5. Turn the grams of Sr into moles (a count of atoms): divide by its atomic weight, 87.62 g/mol

   $\frac{10\text{\hspace{0.17em}g}}{87.62\text{\hspace{0.17em}g/mol}}$$=0.114129\text{\hspace{0.17em}mol}$

6. Divide every mole amount by the smallest (0.114129 mol). That makes the scarcest element 1 and shows the ratio

   $\text{Pd 3.54 : Si 2.433 : Ag 4.386 : Sr 1}$

7. Some ratios aren't whole numbers, so multiply them all by 2. Atoms only come in whole numbers

   $\text{Pd 7.081 : Si 4.867 : Ag 8.773 : Sr 2}$

8. No small multiplier gives near-whole numbers, so this is the closest fit. Check the data

   $\text{Pd 7 : Si 5 : Ag 9 : Sr 2 → Pd₇Si₅Ag₉Sr₂}$

5. Empirical formula: Si₄₅NdSm₁₁Mn₄

1. Turn the grams of Si into moles (a count of atoms): divide by its atomic weight, 28.085 g/mol

   $\frac{50\text{\hspace{0.17em}g}}{28.085\text{\hspace{0.17em}g/mol}}$$=1.78031\text{\hspace{0.17em}mol}$

2. Turn the grams of Nd into moles (a count of atoms): divide by its atomic weight, 144.24 g/mol

   $\frac{5.7\text{\hspace{0.17em}g}}{144.24\text{\hspace{0.17em}g/mol}}$$=0.039517\text{\hspace{0.17em}mol}$

3. Turn the grams of Sm into moles (a count of atoms): divide by its atomic weight, 150.36 g/mol

   $\frac{65\text{\hspace{0.17em}g}}{150.36\text{\hspace{0.17em}g/mol}}$$=0.432296\text{\hspace{0.17em}mol}$

4. Turn the grams of Mn into moles (a count of atoms): divide by its atomic weight, 54.938 g/mol

   $\frac{9\text{\hspace{0.17em}g}}{54.938\text{\hspace{0.17em}g/mol}}$$=0.163821\text{\hspace{0.17em}mol}$

5. Divide every mole amount by the smallest (0.039517 mol). That makes the scarcest element 1 and shows the ratio

   $\text{Si 45.051 : Nd 1 : Sm 10.939 : Mn 4.146}$

6. No small multiplier gives near-whole numbers, so this is the closest fit. Check the data

   $\text{Si 45 : Nd 1 : Sm 11 : Mn 4 → Si₄₅NdSm₁₁Mn₄}$

6. Empirical formula: Tb₁₅Se₃Db₁₁Rb₃

1. Take a 100 g sample, so each percentage becomes that many grams

   $\text{Tb: 55 g, Se: 5.4 g, Db: 68 g, Rb: 6 g}$

2. Turn the grams of Tb into moles (a count of atoms): divide by its atomic weight, 158.93 g/mol

   $\frac{55\text{\hspace{0.17em}g}}{158.93\text{\hspace{0.17em}g/mol}}$$=0.346064\text{\hspace{0.17em}mol}$

3. Turn the grams of Se into moles (a count of atoms): divide by its atomic weight, 78.971 g/mol

   $\frac{5.4\text{\hspace{0.17em}g}}{78.971\text{\hspace{0.17em}g/mol}}$$=0.06838\text{\hspace{0.17em}mol}$

4. Turn the grams of Db into moles (a count of atoms): divide by its atomic weight, 268 g/mol

   $\frac{68\text{\hspace{0.17em}g}}{268\text{\hspace{0.17em}g/mol}}$$=0.253731\text{\hspace{0.17em}mol}$

5. Turn the grams of Rb into moles (a count of atoms): divide by its atomic weight, 85.468 g/mol

   $\frac{6\text{\hspace{0.17em}g}}{85.468\text{\hspace{0.17em}g/mol}}$$=0.070202\text{\hspace{0.17em}mol}$

6. Divide every mole amount by the smallest (0.06838 mol). That makes the scarcest element 1 and shows the ratio

   $\text{Tb 5.061 : Se 1 : Db 3.711 : Rb 1.027}$

7. Some ratios aren't whole numbers, so multiply them all by 3. Atoms only come in whole numbers

   $\text{Tb 15.183 : Se 3 : Db 11.132 : Rb 3.08}$

8. No small multiplier gives near-whole numbers, so this is the closest fit. Check the data

   $\text{Tb 15 : Se 3 : Db 11 : Rb 3 → Tb₁₅Se₃Db₁₁Rb₃}$

7. Empirical formula: Sg₂₂Eu₆Sn₆₃Tl₁₀

1. Take a 100 g sample, so each percentage becomes that many grams

   $\text{Sg: 36 g, Eu: 5.5 g, Sn: 45 g, Tl: 12 g}$

2. Turn the grams of Sg into moles (a count of atoms): divide by its atomic weight, 269 g/mol

   $\frac{36\text{\hspace{0.17em}g}}{269\text{\hspace{0.17em}g/mol}}$$=0.133829\text{\hspace{0.17em}mol}$

3. Turn the grams of Eu into moles (a count of atoms): divide by its atomic weight, 151.96 g/mol

   $\frac{5.5\text{\hspace{0.17em}g}}{151.96\text{\hspace{0.17em}g/mol}}$$=0.036194\text{\hspace{0.17em}mol}$

4. Turn the grams of Sn into moles (a count of atoms): divide by its atomic weight, 118.71 g/mol

   $\frac{45\text{\hspace{0.17em}g}}{118.71\text{\hspace{0.17em}g/mol}}$$=0.379075\text{\hspace{0.17em}mol}$

5. Turn the grams of Tl into moles (a count of atoms): divide by its atomic weight, 204.38 g/mol

   $\frac{12\text{\hspace{0.17em}g}}{204.38\text{\hspace{0.17em}g/mol}}$$=0.058714\text{\hspace{0.17em}mol}$

6. Divide every mole amount by the smallest (0.036194 mol). That makes the scarcest element 1 and shows the ratio

   $\text{Sg 3.698 : Eu 1 : Sn 10.473 : Tl 1.622}$

7. Some ratios aren't whole numbers, so multiply them all by 6. Atoms only come in whole numbers

   $\text{Sg 22.185 : Eu 6 : Sn 62.841 : Tl 9.733}$

8. No small multiplier gives near-whole numbers, so this is the closest fit. Check the data

   $\text{Sg 22 : Eu 6 : Sn 63 : Tl 10 → Sg₂₂Eu₆Sn₆₃Tl₁₀}$

8. Empirical formula: Os₃₄P₄₃O₄₉₄Cm₈

1. Turn the grams of Os into moles (a count of atoms): divide by its atomic weight, 190.23 g/mol

   $\frac{36\text{\hspace{0.17em}g}}{190.23\text{\hspace{0.17em}g/mol}}$$=0.189245\text{\hspace{0.17em}mol}$

2. Turn the grams of P into moles (a count of atoms): divide by its atomic weight, 30.974 g/mol

   $\frac{7.4\text{\hspace{0.17em}g}}{30.974\text{\hspace{0.17em}g/mol}}$$=0.23891\text{\hspace{0.17em}mol}$

3. Turn the grams of O into moles (a count of atoms): divide by its atomic weight, 15.999 g/mol

   $\frac{44\text{\hspace{0.17em}g}}{15.999\text{\hspace{0.17em}g/mol}}$$=2.750172\text{\hspace{0.17em}mol}$

4. Turn the grams of Cm into moles (a count of atoms): divide by its atomic weight, 247 g/mol

   $\frac{11\text{\hspace{0.17em}g}}{247\text{\hspace{0.17em}g/mol}}$$=0.044534\text{\hspace{0.17em}mol}$

5. Divide every mole amount by the smallest (0.044534 mol). That makes the scarcest element 1 and shows the ratio

   $\text{Os 4.249 : P 5.365 : O 61.754 : Cm 1}$

6. Some ratios aren't whole numbers, so multiply them all by 8. Atoms only come in whole numbers

   $\text{Os 33.995 : P 42.917 : O 494.031 : Cm 8}$

7. Round to whole numbers to get the simplest formula

   $\text{Os 34 : P 43 : O 494 : Cm 8 → Os₃₄P₄₃O₄₉₄Cm₈}$

9. Empirical formula: Si₁₅₀Ag₉Hf₂₉Sm₇

1. Turn the grams of Si into moles (a count of atoms): divide by its atomic weight, 28.085 g/mol

   $\frac{36\text{\hspace{0.17em}g}}{28.085\text{\hspace{0.17em}g/mol}}$$=1.281823\text{\hspace{0.17em}mol}$

2. Turn the grams of Ag into moles (a count of atoms): divide by its atomic weight, 107.87 g/mol

   $\frac{8.4\text{\hspace{0.17em}g}}{107.87\text{\hspace{0.17em}g/mol}}$$=0.077872\text{\hspace{0.17em}mol}$

3. Turn the grams of Hf into moles (a count of atoms): divide by its atomic weight, 178.49 g/mol

   $\frac{44\text{\hspace{0.17em}g}}{178.49\text{\hspace{0.17em}g/mol}}$$=0.246512\text{\hspace{0.17em}mol}$

4. Turn the grams of Sm into moles (a count of atoms): divide by its atomic weight, 150.36 g/mol

   $\frac{9\text{\hspace{0.17em}g}}{150.36\text{\hspace{0.17em}g/mol}}$$=0.059856\text{\hspace{0.17em}mol}$

5. Divide every mole amount by the smallest (0.059856 mol). That makes the scarcest element 1 and shows the ratio

   $\text{Si 21.415 : Ag 1.301 : Hf 4.118 : Sm 1}$

6. Some ratios aren't whole numbers, so multiply them all by 7. Atoms only come in whole numbers

   $\text{Si 149.905 : Ag 9.107 : Hf 28.829 : Sm 7}$

7. No small multiplier gives near-whole numbers, so this is the closest fit. Check the data

   $\text{Si 150 : Ag 9 : Hf 29 : Sm 7 → Si₁₅₀Ag₉Hf₂₉Sm₇}$

10. Empirical formula: Ir₁₄Hg₂Fl₁₀He₁₆₇

1. Turn the grams of Ir into moles (a count of atoms): divide by its atomic weight, 192.22 g/mol

   $\frac{45\text{\hspace{0.17em}g}}{192.22\text{\hspace{0.17em}g/mol}}$$=0.234107\text{\hspace{0.17em}mol}$

2. Turn the grams of Hg into moles (a count of atoms): divide by its atomic weight, 200.59 g/mol

   $\frac{6.6\text{\hspace{0.17em}g}}{200.59\text{\hspace{0.17em}g/mol}}$$=0.032903\text{\hspace{0.17em}mol}$

3. Turn the grams of Fl into moles (a count of atoms): divide by its atomic weight, 290 g/mol

   $\frac{47\text{\hspace{0.17em}g}}{290\text{\hspace{0.17em}g/mol}}$$=0.162069\text{\hspace{0.17em}mol}$

4. Turn the grams of He into moles (a count of atoms): divide by its atomic weight, 4.0026 g/mol

   $\frac{11\text{\hspace{0.17em}g}}{4.0026\text{\hspace{0.17em}g/mol}}$$=2.748214\text{\hspace{0.17em}mol}$

5. Divide every mole amount by the smallest (0.032903 mol). That makes the scarcest element 1 and shows the ratio

   $\text{Ir 7.115 : Hg 1 : Fl 4.926 : He 83.525}$

6. Some ratios aren't whole numbers, so multiply them all by 2. Atoms only come in whole numbers

   $\text{Ir 14.23 : Hg 2 : Fl 9.851 : He 167.05}$

7. No small multiplier gives near-whole numbers, so this is the closest fit. Check the data

   $\text{Ir 14 : Hg 2 : Fl 10 : He 167 → Ir₁₄Hg₂Fl₁₀He₁₆₇}$