Question

0.545 g of aluminum burns completely in oxygen, producing 1.030 g of aluminum oxide. find the empirical formula of the oxide​

Answer 1

Answer:

1.030-0.545=0.485 grams of oxygen in compound.

Convert grams of Al and grams of O into moles:

0.545g Al x 1 mol Al/26.98g Al = 0.0202 mol Al

0.485g O x 1 mol O/16.00g O = 0.0303 mol O

Next we divide the moles of both of them by the lowest number to get simplest ratio

0.0202/0.0202 = 1 Al

0.0303/0.0202 = 1.5 O

Next we multiply all numbers until we get a whole number for each one. In this case 2 works:

Al 1 x 2 : O 1.5 x 2

and we are left with Al2O3

Explanation:

Hope this helps :D

Answer 2

Answer:

mass Al = 0.545 g

\text{mass O = 0.485 g}mass O = 0.485 g

Required:

empirical formula

Solution:

Step 1: Calculate the number of moles of each element

n \: \text{Al = 0.545 g} \times \frac{\text{1 mol}}{\text{26.98 g}} = \text{0.0202 mol}nAl = 0.545 g×

26.98 g

1 mol

=0.0202 mol

n \: \text{O = 0.485 g} \times \frac{\text{1 mol}}{\text{16.00 g}} = \text{0.0303 mol}nO = 0.485 g×

16.00 g

1 mol

=0.0303 mol

Step 2: Represent an empirical formula

empirical \: formula = \text{Al}_{x}\text{O}_{y}empiricalformula=Al

x

O

y

Step 3: Divide the number of moles of each element by the least number of moles

x = \frac{\text{0.0202 mol}}{\text{0.0202 mol}} = 1 × 2 = 2x=

0.0202 mol

0.0202 mol

=1×2=2

y = \frac{\text{0.0303 mol}}{\text{0.0202 mol}} = 1.5 × 2 = 3y=

0.0202 mol

0.0303 mol

=1.5×2=3

Step 4: Write the empirical formula

\blue{empirical \: formula = \text{Al}_{2}\text{O}_{3}}empiricalformula=Al

2

O

3

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