# Theoretical Yield Calculator

Moles

## Introduction of Theoretical Yield Calculator

Are you facing problems regarding chemistry reaction products calculations? We are here to discuss and to solve these issues of the chemistry lab. The theoretical yield calculator is useful to calculate the weight of reactants you are about using into the process, look up mole ratio and molar weight then enter them into the calculator. It will let you know how many grams product generates in this reaction. This calculator is very easy to use. It just takes yield equation and solves it and gives theoretical yield in the result.

## Theoretical Yield Formula

To determine the theoretical yield of any chemical reaction, multiply the number of moles by the molecular weight. Theoretical yield will be calculated in grams because it is the amount of the expected product. Now we will solve example to make it more clear. Before moving towards example, I feel it necessary to share commonly used terms relevant to our todays’ topic.

## Terminologies

### What is the yield?

It is the amount of any product.

## What is the theoretical yield??

It is the maximum amount of product that can be obtained in a chemical reaction (calculated from the limiting reagent). So the theoretical yield is something that almost never obtained because theoretical yield can be obtained under the perfect conditions. The perfect conditions mean where no product is lost in the process, means; no impurities mixed in the reactants, no production of unexpected byproducts and no loss of product because of measurement etc.

### What is an actual yield?

It is an amount of product actually obtained from a chemical reaction (must be experimentally determined, cannot be calculated – never more than the theoretical yield).

### What is limiting reagent?

The limiting reagent is the reagent that will be used up completely, thus limiting the extent of the reaction.

### What is the percent yield?

Percent yield is the ratio of actual yield to theoretical yield.

$$\text{Percent Yield}\;=\;\frac{actual\;yield}{theoretical\;yield}\;\;* 100%$$

## How to Calculate Theoretical Yield?

Here, I am going to elaborate how to calculate theoretical yield step by step. There are a few steps; by following them we can calculate how many grams of product each reagent can produce.

Step 1: Chemical equations must be balanced equations

Step 2: Determine the mole ratio between the reactants and the products

Step 3: Now calculate the theoretical yield by the help of the above information.

### Example:

3.45g of CH3Br is reacted with 5.23g NaOH

Calculate the theoretical yield of NaBr according to the following balanced chemical equation.

$$CH_3Br\;+\;NaOH\;\rightarrow\;CH_3OH\;+\;NaBr$$

We need to figure out how many moles of each reagent are present so that we can find the limiting reagent.

First, we will convert the reagents listed from grams to moles. Next, we will find how many moles of the product we can form with our calculated moles of reagent. We can do this all just in one step with a little dimensional analysis.

$$3.45g\;\;CH_3Br\;*\;\frac{1\;mole\;CH_3Br}{94.94g\;CH_3Br}\;*\;\frac{1\;mole\;NaBr}{1\;mole\;CH_3Br}$$

$$=0.0363\;\text{moles}\;NaBr$$

Now we will do the same for sodium hydroxide.

$$5.23g\;\;NaOH\;*\;\frac{1\;mole\;NaOH}{39.99g\;NaOH}\;*\;\frac{1\;mole\;NaBr}{1\;mole\;NaOH}$$

$$=0.1308\;\text{moles}\;NaBr$$

$$0.0363\;\;<\;\;0.1308$$

We can see that the 3.45 grams of methyl bromide is going to produce less moles of bromine than hydroxide and thus, all of the methyl bromide will react completely. Therefore, methyl bromide is our limiting reagent.

This theoretical yield tells us how much product we should end up with if our reaction runs to completion. So, after using up all of the methyl bromide, we are left with 0.0363 moles of bromine; this is our theoretical yield.

We can use this theoretical yield when performing an experiment, along with the measured actual yield, to calculate percentage yield.

Now, let’s figure out what the theoretical yield would be in terms of grams.

$$0.0363\;\text{moles}\;NaBr\;*\;39.99\;\frac{g}{mol}\;=\;1.452g\;NaBr$$

We see there is expected to be 1.452 grams of sodium bromide product.

To calculate percent yield, you simply take actual yield 1.099 grams of sodium bromide, divided by the theoretical yield 1.452 grams of sodium bromide.

$$\frac{1.099g\;NaBr}{1.452g\;NaBr}\;*\;100\;=\;75.69% yield$$