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olya-2409 [2.1K]
2 years ago
10

A sample of an alloy of aluminum contains 0.0898 mol Al and 0.0381 mol Mg. What are the mass percentages of Al and Mg in the all

oy?
Chemistry
2 answers:
blondinia [14]2 years ago
7 0

Answer:

Al 72.61%

Mg 27.39%

Explanation:

To obtain the mass percentages, we need to place the individual masses over the total mass and multiply by 100%.

If we observe clearly, we can see that the parameters given are the moles. We need to convert the moles to mass.

To do this ,we need to multiply the moles by the atomic masses. The atomic mass of aluminum is 27 while that of magnesium is 24.

Now, the mass of aluminum is thus = 27 * 0.0898 = 2.4246g

The mass of magnesium is 0.0381 * 24 = 0.9144g

We can now calculate the mass percentage.

The total mass is 0.9144 + 2.4246 = 3.339g

% mass of Al = 2.4246/3.339 * 100 = 72.61%

% mass of Mg = 0.9144/3.39 * 100 = 27.39%

marshall27 [118]2 years ago
4 0

Answer:

Mass % Al = 72.3 %

Mass % Mg = 27.7 %

Explanation:

Step 1: Data given

Number of moles Al = 0.0898 moles

Number of moles Mg = 0.0381 moles

Molar mass Al = 26.98 g/mol

Molar mass Mg = 24.3 g/mol

Step 2: Calculate mass Al

Mass Al = moles Al * molar mass Al

Mass Al = 0.0898 moles * 26.98 g/mol

Mass Al = 2.42 grams

Step 3: Calculate mass Mg

Mass Mg = 0.0381 moles * 24.3 g/mol

Mass Mg = 0.926 grams

Step 4: Calculate total mass

Total mass = mass Al + mass Mg

Total mass = 2.42 grams + 0.926 grams

Total mass = 3.346 grams

Step 5: Calculate mass %

Mass % Al = (mass Al/ total mass) * 100%

Mass % Al = (2.42 grams / 3.346 grams ) *100%

Mass % Al = 72.3 %

Mass % Mg = (mass Mg/ total mass)*100%

Mass % Mg = (0.926 / 3.346) *100 %

Mass % Mg = 27.7 %

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To formulate the empirical formula, we need to follow some steps:

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Hence, the empirical and molecular formula for the given organic compound is C_9H_{12}O and C_{18}H_{24}O_2

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