# Solving a Chemistry Equation: Question 10

• Chemistry
• mutineer123
In summary, the conversation is about solving question 10 of a chemistry assignment, specifically finding the values of x and y in a set of equations. The first equation was found by using the atomic masses of all the elements in the formula, while the second equation involves the empirical formula. The conversation also discusses finding the charge on the resulting ion and calculating percentages. The speaker ultimately solves the problem with the help of another person.
mutineer123

## Homework Statement

http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_s11_qp_12.pdf

question 10

## The Attempt at a Solution

The way I am doing it, is to get two equations in terms of x and y and then equate them. For the first equation I used the fact that all of the element's individual atomic masses will add up to 571.5. So i got 27x + 28.1y = 172
My problem now is how to get the second equation? I think it has something to do with the empirical formula, but I can't figure it out. Help!

If you remove the Al and Si from the formula, what is the charge on the resulting ion?

What is 14.04% of 571.5?

Borek said:
What is 14.04% of 571.5?

I think I got it, thanks to you. So for x we find 14.17% of 571.5. That answer we divide by 27, (because we know that the mole of aluminium is 1), and the x i got is 2.999(pretty close to one of the options). Same for y. Thanks!

I would suggest approaching this problem by first identifying the elements involved in the equation and their respective atomic masses. Then, using the given information that the sum of the atomic masses is 571.5, you can set up an equation in terms of x and y for each element. For example, for element X, the equation would be 27x = 172 - 28.1y.

Next, you can use the given information about the empirical formula to determine the ratio between x and y. The empirical formula states that the ratio of elements in a compound is always in its simplest form. In this case, the ratio between x and y is 3:1, so you can substitute this ratio into your equations to solve for x and y.

Once you have solved for x and y, you can then plug these values into the empirical formula to determine the molecular formula for the compound. Remember to check your solution by ensuring that the atomic masses of each element in the molecular formula add up to the given sum of 571.5. I hope this helps and good luck with your homework!

## 1. How do I balance a chemistry equation?

To balance a chemistry equation, you need to make sure that the number of atoms of each element on the reactant side is equal to the number of atoms on the product side. You can do this by adjusting the coefficients in front of each compound.

## 2. What is the purpose of balancing a chemistry equation?

The purpose of balancing a chemistry equation is to ensure that the law of conservation of mass is followed. This means that the total mass of the reactants must be equal to the total mass of the products.

## 3. What are the steps for balancing a chemistry equation?

The steps for balancing a chemistry equation are as follows:

1. Identify the reactants and products
2. Count the number of atoms of each element in the reactants and products
3. Balance the elements one at a time by adjusting the coefficients
4. Check that the number of atoms of each element is equal on both sides
5. If the equation is not balanced, continue adjusting the coefficients until it is

## 4. Can you give an example of balancing a chemistry equation?

For example, let's balance the equation: H2 + O2 → H2O

First, we count the number of atoms of each element: 2 hydrogen atoms and 2 oxygen atoms on the reactant side, and 2 hydrogen atoms and 1 oxygen atom on the product side.

To balance the hydrogen, we need to put a coefficient of 2 in front of the water molecule: H2 + O2 → 2H2O

Now, we have 4 oxygen atoms on the product side, so we need to balance the oxygen by putting a coefficient of 2 in front of the oxygen molecule: 2H2 + 2O2 → 2H2O

Finally, we have 4 hydrogen atoms and 4 oxygen atoms on both sides, and the equation is balanced.

## 5. What do I do if I encounter fractions while balancing a chemistry equation?

If you encounter fractions while balancing a chemistry equation, you can multiply the entire equation by the lowest common multiple (LCM) of all the denominators to get rid of the fractions. Then, continue balancing the equation as usual.

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