# 6.25 grams of pure iron are allowed to react with oxygen to form an oxide. If the product weighs 14.31 grams, find the simplest formula of the compound?

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If you could, can you go through the process? I got 2 Iron per 9 Oxygen but I don't know if it's right.

Also, in determining the simplest formula of lead sulfide, 2.46 grams of lead are place in a crucible with 2.00 grams of sulfur. When the reaction is complete, the product has a mass of 3.22 grams. What mass of sulfur should be used in the simplest formula calculation? Find the simplest formula of lead sulfide.

Thanks!

If you could, can you go through the process? I got 2 Iron per 9 Oxygen but I don't know if it's right.

Also, in determining the simplest formula of lead sulfide, 2.46 grams of lead are place in a crucible with 2.00 grams of sulfur. When the reaction is complete, the product has a mass of 3.22 grams. What mass of sulfur should be used in the simplest formula calculation? Find the simplest formula of lead sulfide.

Thanks!

##### 1 Answer

Here's what I got.

#### Explanation:

Your goal when determining the *simplest formula*, which is another term used for **empirical formula**, is to find how many **moles** of each element you have in your sample.

This will then allow you to find the **mole ratio** that exist between the elements that make up the compound.

So, you are told that a

Right from the start, you should know that, according to the **law of mass conservation**, the mass of the product **must** be equal to the combined masses of the two reactants.

#m_"oxide" = m_(O) + m_(Fe)#

This means that your sample will contain

#m_(O) = "14.31 g" - "6.25 g" = "8.06 g O"#

Now that you know how many grams of each element you have in the sample, use their respective **molar masses** to find how many *moles* of each you have

#6.25 color(red)(cancel(color(black)("g"))) * "1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = "0.1119 moles Fe"#

#8.06 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "0.5038 moles O"#

Now, to get the mole ratio that exists between the two elements in the compound, divide both values by the *smallest one*. This will get you

#"For Fe; " (0.1119 color(red)(cancel(color(black)("moles"))))/(0.1119color(red)(cancel(color(black)("moles")))) = 1#

#"For O: " (0.5038color(red)(cancel(color(black)("moles"))))/(0.1119color(red)(cancel(color(black)("moles")))) = 4.5#

The **empirical formula** for this compound **must** contain the **smallest whole number ratio** that exists between the elements. Since you have a **whole number** ratio that matches this one is

This means that the compound's empirical formula will be

#("Fe"_1"O"_4.5)_2 implies color(green)("Fe"_2"O"_9)#

I'll leave the second problem to you as practice. The **exact** same approach applies here as well.

The mass of sulfur **must** account for the difference between the mass of the sulfide and the mass of lead *mass conservation*.

From that point on, use the molar masses of the two elements to get the number of moles, then divide both values by the smallest one to get the mole ratio.

Your answer should be