Question

The sample calculation for iron oxide in the IDEAS section of this experiment used known atomic weights to calculate an empirical formula. However, early chemists did not have any references in which they could look up atomic weights. Instead, they guessed at the formulas of compounds and measured the percent compositions of elements in compounds in order to calculate atomic weights. Calculate an atomic weight for iron using the hypothetical formula Fe101 and the composition data given in the example in the IDEAS section. You may assume the atomic weight of oxygen is known from other sources to be 16 amu.

Answer 1

37.33 grams

Step-by-step explanation:

The missing information embedded in the idea section is attached in the image below:

The aim of this question is to determine the atomic wt. of Iron (Fe) from the hypothetic formula:

Fe₁O₁

Here, we know that the mole ratio can be written as:

`(O)/(Fe)=(1)/(1)`

Suppose we assume that the atomic wt. of Fe = β(unknown)???

Then the grams of O and Fe that is contained in Fe₁O₁ can be expressed as:

For O:

1 × 16 grams of Oxygen = 16 grams of O

For Fe:

1 × β grams of Fe = β grams of Fe

Now, let's take a look at the idea experiment, the mole solution can be computed as:

`(O)/(Fe) = (3)/(2) \\ \\ \text{It implies that} \implies \frac{(3* 16) \text{grams of O}}{(2 * 56 ) \ \text{grams of Fe}}`

Equating both expressions above, we have:

`\implies (16)/( \beta) = (3* 16)/(2* 56)`

`{ \beta} = ((2* 56)* 16)/( 3* 16)`

`\mathbf{{ \beta} = 37.33 \ grams}`