L39 .. I don't understand it. May someone please explain it to me?! Thank You.

Moderators: Chem_Mod, Chem_Admin

JasonNovik3A
Posts: 21
Joined: Fri Sep 29, 2017 7:07 am

L39 .. I don't understand it. May someone please explain it to me?! Thank You.

Postby JasonNovik3A » Thu Oct 05, 2017 8:29 pm

In simple terms haha

Golbarg Rahimi 3k
Posts: 26
Joined: Sat Jul 22, 2017 3:01 am

Re: L39 .. I don't understand it. May someone please explain it to me?! Thank You.

Postby Golbarg Rahimi 3k » Thu Oct 05, 2017 9:00 pm

By subtracting 26.45 from 28.35 ( 28.35-26.45=1.90) you will have the actual grams of product formed. you already know that the product is made of tin ( Sn) and Oxygen. with the initial amount of tin available ( 1.50g) you can determine the moles of tin available and produced. we have 1.90 grams of product and 1.50 grams of it ,is the mass of tin therefore ( 1.90-1.50) grams is the mass of oxygen. by using the mass of oxygen and the molar mass you determine the moles of oxygen available in the product. Next with the molar ratios, Sn (1) and O (2) you can write the empirical formula : SnO2 which is tin (IV) oxide

Leon Popa
Posts: 18
Joined: Fri Sep 29, 2017 7:04 am

Re: L39 .. I don't understand it. May someone please explain it to me?! Thank You.

Postby Leon Popa » Thu Oct 05, 2017 9:30 pm

Hey Jason,

Here's how I approach the problem:

We have 1.50 grams of Tin, which is abbreviated "Sn" on the periodic table. They tell us the 1.50 g of Sn is placed in a crucible (a metal container) that has capacity of 26.45 grams. We also know that the Tin reacts in this container with oxygen in the air to form an oxide. If the oxygen is in the air, that must mean its in the form of O2. To represent this reaction, I wrote on my paper:

Sn + O2 -> "Oxide"

Apparently the container and the oxide together weigh 28.35 g. We know the container by itself weighs 26.45 g. So 28.35 g - 26.45 g gives us the weight of the oxide, 1.90 g.

This is fine, but lets take a moment to understand what we're ultimately looking for. We want the empirical formula of this unnamed oxide. To do that we need the moles of the reactants, and to get the moles, we need the weight of the reactants.

1.50 g of Sn reacts with some weight of O2 to produce 1.90 g of oxide. 1.90 g - 1.50 g gives us the weight of O2, 0.40 g. We can be sure of this because the reaction must conserve mass, so the reactants together must weigh as much as the product.

So we now have 1.50g Sn and 0.40 O2. To calculate the moles:

1.50 g Sn / 118.71 g/mol = 0.0126 mol Sn

0.40 g O2 / 2*16 g/mol = 0.0125 mol O2

So far we have a ratio of Sn to O2 of 0.0126 : 0.0125. Conveniently enough, the moles of Sn and O2 are practically the same. So we can say the ratio of Sn to O2 is 1:1.

This means that the ratio of Sn to O2 in the product must also be 1:1. Therefore, in our oxide, we have 1 mole of Sn and 1 mole of O2.

The empirical formula is SnO2.

As for the name of SnO2, break it up into parts. Part 1 is Sn, that's just "Tin." Part 2 is O2. The problem already told us the compound is an "oxide". So Part 2 must be that "oxide" part.

The name is Tin Oxide.

Hope that helps Jason.

- Leon


Return to “Limiting Reactant Calculations”

Who is online

Users browsing this forum: No registered users and 1 guest