Question

1. the percentage by mass of br in the compound $albr_3$ is closest to

a. 10.% b. 25% c. 75% d. 90.%

1. the percent composition by mass of magnesium in $mgbr_2$ (gram-formula mass = 184 grams/mole) is equal to

a. $\frac{24}{184} \times 100$ b. $\frac{160.}{184} \times 100$
c. $\frac{184}{24} \times 100$ d. $\frac{184}{160.} \times 100$

1. what is the percent composition by mass of nitrogen in $nh_4no_3$ (gram-formula mass = 80.0 grams/mole)?

a. 17.5% b. 35.0% c. 52.5% d. 60.0%

1. what is the percent composition by mass of aluminum in $al_2(so_4)_3$ (gram-formula mass = 342 grams/mole)?

a. 7.89% b. 15.8% c. 20.8% d. 36.0%

1. what is the percent by mass of oxygen in propanal, $ch_3ch_2cho$?

a. 10.0% b. 27.6% c. 38.1% d. 62.1%

1. in which compound is the percent by mass of oxygen greatest?

a. $beo$ b. $mgo$ c. $cao$ d. $sro$

1. what is the percent by mass of oxygen in $ca(oh)_2$? formula mass = 74.1

a. 21.6% b. 43.2%
c. 45.9% d. 54.1%

Explanation:

Step1: Calculate molar mass of AlBr₃

Molar mass: $M(\text{Al}) + 3\times M(\text{Br}) = 27 + 3\times79.9 = 266.7$ g/mol

Step2: Find mass % of Br

$\text{Mass \% Br} = \frac{3\times79.9}{266.7} \times 100\% \approx 90\%$
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Step1: Identify Mg's molar mass

$M(\text{Mg}) = 24$ g/mol

Step2: Set up mass % formula

$\text{Mass \% Mg} = \frac{M(\text{Mg})}{\text{Gram-formula mass of MgBr}_2} \times 100 = \frac{24}{184} \times 100$
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Step1: Count N atoms in NH₄NO₃

2 nitrogen atoms per molecule

Step2: Calculate total mass of N

$2\times M(\text{N}) = 2\times14 = 28$ g/mol

Step3: Find mass % of N

$\text{Mass \% N} = \frac{28}{80.0} \times 100\% = 35.0\%$
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Step1: Count Al atoms in Al₂(SO₄)₃

2 aluminum atoms per molecule

Step2: Calculate total mass of Al

$2\times M(\text{Al}) = 2\times27 = 54$ g/mol

Step3: Find mass % of Al

$\text{Mass \% Al} = \frac{54}{342} \times 100\% \approx 15.8\%$
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Step1: Calculate molar mass of propanal

$M(\text{CH}_3\text{CH}_2\text{CHO}) = 3\times12 + 6\times1 + 16 = 58$ g/mol

Step2: Find mass % of O

$\text{Mass \% O} = \frac{16}{58} \times 100\% \approx 27.6\%$
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Step1: Compare mass % of O in oxides

For oxides $\text{MO}$, $\text{Mass \% O} = \frac{16}{M(\text{M})+16} \times 100\%$. Smaller $M(\text{M})$ gives higher %: $M(\text{Be})=9$ (smallest)
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Step1: Count O atoms in Ca(OH)₂

2 oxygen atoms per molecule

Step2: Calculate total mass of O

$2\times M(\text{O}) = 2\times16 = 32$ g/mol

Step3: Find mass % of O

$\text{Mass \% O} = \frac{32}{74.1} \times 100\% \approx 43.2\%$

Answer:

1. D. 90.%
2. A. $\frac{24}{184} \times 100$
3. B. 35.0%
4. B. 15.8%
5. B. 27.6%
6. A. BeO
7. B. 43.2%