Question

1.08 g of Aluminium combines with 0.96 g of oxygen to form an oxide. What is the
empirical formula of the oxide? [relative atomic mass: O=16, Al=27]​

Answer 1

Answer: The empirical formula is $Al_2O_3$

Explanation:

The empirical formula of aluminum oxide, which has 1.08 g of aluminum combines chemically with 0. 96 g of oxygen. Since the ratio must contain the simplest whole number the ratio is 2:3, thus the empirical formula is $Al_2O_3$

Step 1: List the given values.

Mass of Al = 1.08g

Mass of O = 0.96

Step 2: Calculate the number of moles of each element.

$Al = 1.08 * \frac{1}{26.98} = 0.0400\\ O = 0.96 *\frac{1}{16}= 0.06$

Step 3: Represent an empirical formula.

$Al_xO_y$

Step 4: Divide the number of moles of each element by the least number of moles to get the subscripts of each element.

Since the smallest number of moles is 0.0404 mol, we divide each number of moles by 0.0404 mol.

$x = \frac{0.0400}{0.0404} = 0.99 = 1 (approx.)\\ y = \frac{0.0600}{0.0404} = 1.485 = 1.5(approx)$

Step 5: Multiply the subscripts obtained by a certain number to become a whole number.

x = 1*2 = 2

y = 1.5*2 = 3

Since the decimal part in 1.5 is 0.5, we need to multiply all the subscripts by 2. Therefore,

Step 6: Write the empirical formula.

$Al_2O_3$

Hence, the empirical formula is Al₂O₃.

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