Question

find the molar masses of the following compounds and circle your answer. remember to show your work and include your units and one decimal place1) nabr2) pbso₄3) ca(oh)₂4) na₃po₄5) (nh₄)₂co₃6) c₆h₁₂o₆7) fe₃(po₄)₂8) (nh₄)₂s9) zn(c₂h₃o₂)₂10) na₃b₄o₇·10h₂o

Explanation:

1) NaBr

Step1: Sumar masas atómicas

$\text{Masa molar} = \text{Masa Na} + \text{Masa Br} = 22.0\ \text{g/mol} + 79.9\ \text{g/mol}$

Step2: Calcular total

$\text{Masa molar} = 101.9\ \text{g/mol}$

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2) $\text{PbSO}_4$

Step1: Sumar masas atómicas

$\text{Masa molar} = \text{Masa Pb} + \text{Masa S} + 4\times\text{Masa O} = 207.2\ \text{g/mol} + 32.1\ \text{g/mol} + 4\times16.0\ \text{g/mol}$

Step2: Calcular total

$\text{Masa molar} = 207.2 + 32.1 + 64.0 = 303.3\ \text{g/mol}$

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3) $\text{Ca(OH)}_2$

Step1: Sumar masas atómicas

$\text{Masa molar} = \text{Masa Ca} + 2\times\text{Masa O} + 2\times\text{Masa H} = 40.1\ \text{g/mol} + 2\times16.0\ \text{g/mol} + 2\times1.0\ \text{g/mol}$

Step2: Calcular total

$\text{Masa molar} = 40.1 + 32.0 + 2.0 = 74.1\ \text{g/mol}$

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4) $\text{Na}_3\text{PO}_4$

Step1: Sumar masas atómicas

$\text{Masa molar} = 3\times\text{Masa Na} + \text{Masa P} + 4\times\text{Masa O} = 3\times22.0\ \text{g/mol} + 31.0\ \text{g/mol} + 4\times16.0\ \text{g/mol}$

Step2: Calcular total

$\text{Masa molar} = 66.0 + 31.0 + 64.0 = 161.0\ \text{g/mol}$

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5) $(\text{NH}_4)_2\text{CO}_3$

Step1: Sumar masas atómicas

$\text{Masa molar} = 2\times\text{Masa N} + 8\times\text{Masa H} + \text{Masa C} + 3\times\text{Masa O} = 2\times14.0\ \text{g/mol} + 8\times1.0\ \text{g/mol} + 12.0\ \text{g/mol} + 3\times16.0\ \text{g/mol}$

Step2: Calcular total

$\text{Masa molar} = 28.0 + 8.0 + 12.0 + 48.0 = 96.0\ \text{g/mol}$

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6) $\text{C}_6\text{H}_{12}\text{O}_6$

Step1: Sumar masas atómicas

$\text{Masa molar} = 6\times\text{Masa C} + 12\times\text{Masa H} + 6\times\text{Masa O} = 6\times12.0\ \text{g/mol} + 12\times1.0\ \text{g/mol} + 6\times16.0\ \text{g/mol}$

Step2: Calcular total

$\text{Masa molar} = 72.0 + 12.0 + 96.0 = 180.0\ \text{g/mol}$

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7) $\text{Fe}_3(\text{PO}_4)_2$

Step1: Sumar masas atómicas

$\text{Masa molar} = 3\times\text{Masa Fe} + 2\times\text{Masa P} + 8\times\text{Masa O} = 3\times55.8\ \text{g/mol} + 2\times31.0\ \text{g/mol} + 8\times16.0\ \text{g/mol}$

Step2: Calcular total

$\text{Masa molar} = 167.4 + 62.0 + 128.0 = 357.4\ \text{g/mol}$

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8) $(\text{NH}_4)_2\text{S}$

Step1: Sumar masas atómicas

$\text{Masa molar} = 2\times\text{Masa N} + 8\times\text{Masa H} + \text{Masa S} = 2\times14.0\ \text{g/mol} + 8\times1.0\ \text{g/mol} + 32.1\ \text{g/mol}$

Step2: Calcular total

$\text{Masa molar} = 28.0 + 8.0 + 32.1 = 68.1\ \text{g/mol}$

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9) $\text{Zn(C}_2\text{H}_3\text{O}_2\text{)}_2$

Step1: Sumar masas atómicas

$\text{Masa molar} = \text{Masa Zn} + 4\times\text{Masa C} + 6\times\text{Masa H} + 4\times\text{Masa O} = 65.4\ \text{g/mol} + 4\times12.0\ \text{g/mol} + 6\times1.0\ \text{g/mol} + 4\times16.0\ \text{g/mol}$

Step2: Calcular total

$\text{Masa molar} = 65.4 + 48.0 + 6.0 + 64.0 = 183.4\ \text{g/mol}$

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10) $\text{Na}_3\text{B}_4\text{O}_7\cdot10\text{H}_2\text{O}$

Step1: Sumar masas atómicas

$\text{Masa molar} = 3\times\text{Masa Na} + 4\times\text{Masa B} + 7\times\text{Masa O} + 10\times(2\times\text{Masa H} + \text{Masa O}) = 3\times22.0\ \text{g/mol} + 4\times10.8\ \text{g/mol} + 7\times16.0\ \text{g/mol} + 10\times(2\times1.0 + 16.0)\ \text{g/mol}$

Step2: Calcular total

$\text{Masa molar} = 66.0 + 43.2 + 112.0 + 10\times18.0 = 66.0 + 43.2 + 112.0 + 180.0 = 401.2\ \text{g/mol}$

Answer:

1. $\text{NaBr}: 101.9\ \text{g/mol}$
2. $\text{PbSO}_4: 303.3\ \text{g/mol}$
3. $\text{Ca(OH)}_2: 74.1\ \text{g/mol}$
4. $\text{Na}_3\text{PO}_4: 161.0\ \text{g/mol}$
5. $(\text{NH}_4)_2\text{CO}_3: 96.0\ \text{g/mol}$
6. $\text{C}_6\text{H}_{12}\text{O}_6: 180.0\ \text{g/mol}$
7. $\text{Fe}_3(\text{PO}_4)_2: 357.4\ \text{g/mol}$
8. $(\text{NH}_4)_2\text{S}: 68.1\ \text{g/mol}$
9. $\text{Zn(C}_2\text{H}_3\text{O}_2\text{)}_2: 183.4\ \text{g/mol}$
10. $\text{Na}_3\text{B}_4\text{O}_7\cdot10\text{H}_2\text{O}: 401.2\ \text{g/mol}$