Główna zawartość

## MCAT

### Kurs: MCAT > Jednostka 8

Lekcja 28: Stechiometria- Stoichiometry questions
- Stoichiometry article
- Stoichiometry and empirical formulae
- Empirical formula from mass composition edited
- Wyznaczanie wzoru empirycznego i rzeczywistego związku w oparciu o skład procentowy
- Mol i liczba Avogadro
- Przykładowy problem stechiometryczny 1
- Stechiometria
- Stechiometria: Reagenty ograniczające
- Limiting reactant example problem 1 edited
- Gęstość względna

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# Stoichiometry article

In our day to day life, it’s child’s play for us to ‘count’ things or ‘weigh’ things. For example, you can easily count and tell me that there are 4 bananas or there are 6 apples kept on the table. Shopkeepers can easily weigh things on a weighing balance and tell us ‘this packet of rice weighs 2 pounds’ or ‘this packet of tea weighs 500 g’. But things change when you start dealing with atoms and molecules. You cannot simply count or weigh atoms or molecules, as you can do with the common items that you encounter in your daily life. Here it gets a little tricky and something termed as ‘Stoichiometry’ kicks in.

## How do you define stoichiometry?

Stoichiometry is the branch of chemistry that deals with the relationship between the relative quantities of substances taking part in a chemical reaction

Let’s write a general chemical reaction where there are two reactants, A and B that react together to form two products, C and D respectively

A + B → C + D

The concept of stoichiometry will help us answer questions like

- if ‘x’ grams of A is present, then how many grams of C or D or both will be produced
- if ‘y’ grams of B is present, then how many grams of C or D or both will be produced
- if ‘x’ grams of A is present and ‘y’ grams of B is present, which is the limiting reactant and which is the reactant present in excess
- if ‘x’ grams of A react with ‘y’ grams of B, how much of B will remain unreacted (or unconsumed) after the reaction is over
- how many grams of A will produce ‘z’ grams of C

The concept of stoichiometry can be best understood by solving real problems involving chemical reactions, reactants and products

Just two things to remember before we get started:

- the first step in any stoichiometric problem is to
*always*ensure that the chemical reaction you are dealing with is balanced, - clarity of the concept of a ‘mole’ and the relationship between ‘amount (grams)’ and ‘moles’.

## ‘Mole concept’ in brief

Let’s write a simple balanced chemical reaction, where hydrogen gas reacts with oxygen gas to produce water.

start color #e84d39, 2, end color #e84d39Hstart subscript, 2, end subscript + Ostart subscript, 2, end subscript → start color #e84d39, 2, end color #e84d39Hstart subscript, 2, end subscriptO

In the above reaction the prefixes (shown in red) refer to the moles of the molecules involved in the chemical reaction. These are also referred to as coefficients. Here, two moles of hydrogen gas are reacting with one mole of oxygen gas to produce two moles of water. ‘Mole’ simply refers to an enormously big number which is 6.023 X 10start superscript, 23, end superscript.

Let’s see how a particular element is represented in a periodic table. Let’s pick chlorine (Cl).

As you can see, there are three pieces of information provided here.

- Atomic symbol of chlorine
- Atomic number of chlorine; this corresponds to the number of protons or electrons present in an atom of chlorine
- Atomic mass of chlorine; this value represents the total mass of both the protons and neutrons present in an atom of chlorine. Atomic mass is commonly referred to as A, or atomic mass number.

Rule of thumb is:

*atomic mass of an element (in grams) = mass of 1 mole of atoms of that element*. So in the case of chlorine (Cl), one mole of chlorine atoms (i.e. 6.023 X 10start superscript, 23, end superscript atoms of chlorine) will weigh 35.453 g.The same principle can be applied to a molecule also, example water (Hstart subscript, 2, end subscriptO).

First we need to calculate the molar mass of Hstart subscript, 2, end subscriptO = 2 (atomic mass of H) + 1(atomic mass of O) = 2(1.008) + 1(15.999) = 18.015 g

**So, one mole of water molecules will weigh 18.015 g**

The conversion between moles and mass (in grams) should be very clear to us before we move on because this is going to be the basis of all stoichiometric calculations.

*Let’s find out how many moles of glucose are present in 25 g of glucose*

The first step, as always, is to calculate the molar mass of glucose (Cstart subscript, 6, end subscriptHstart subscript, 12, end subscriptOstart subscript, 6, end subscript) = 6 (atomic mass of C) + 12 (atomic mass of H) + 6 (atomic mass of O) = 6 (12.011) + 12 (1.008) + 6 (15.999) = 72.066 + 12.096 + 95.994 = 180.156 grams

**So, 1 mole of glucose weighs 180.156 g**

i.e. 180.156 g of glucose = 1 mole of glucose molecules

therefore, 1 g of glucose = (1 mole/180.156 g) x 1 g

so, 25 g of glucose = (1 mole/180.156 g) x 25 g = 0.139 moles

Now let’s practice some stoichiometric problems

### Zadanie 1

If 138.6 g of KClOstart subscript, 3, end subscript is heated and decomposes completely, how many grams of oxygen gas would be produced?

**Step 1: Ensure that the reaction is balanced**

In this case the reaction is not balanced. After balancing, the equation looks like

**Krok 2: Convert grams of KClOstart subscript, 3, end subscript to moles**

For this we first need to calculate the molar mass of KClOstart subscript, 3, end subscript = 1 (atomic mass of K) + 1 (atomic mass of Cl) + 3 (atomic mass of O) = 1 (39.098) + 1 (35.453) + 3 (15.999) = 39.098 + 35.453 + 47.997 = 122.548 g

i.e. 122.548 g of KClOstart subscript, 3, end subscript = 1 mole of KClOstart subscript, 3, end subscript

therefore, 1 g of KClOstart subscript, 3, end subscript = (1 mole/ 122.548 g) x 1 g

so, 138.6 g of KClOstart subscript, 3, end subscript = (1 mole/ 122.548 g) x 138.6 g = 1.130 moles

**Step 3: Use mole ratios (from the balanced chemical reaction) to calculate the moles of the relevant reactant or product**

In this example, we need to calculate the mass of Ostart subscript, 2, end subscript produced. So we first need to calculate the moles of Ostart subscript, 2, end subscript that would be produced from 1.130 moles of KClOstart subscript, 3, end subscript

Looking at the balanced chemical reaction, we can infer that 2 moles of KClOstart subscript, 3, end subscript produce 3 moles of Ostart subscript, 2, end subscript

So, 1.130 moles of KClOstart subscript, 3, end subscript will produce [(3 moles of Ostart subscript, 2, end subscript/ 2 moles of KClOstart subscript, 3, end subscript) x 1.130 moles of KClOstart subscript, 3, end subscript] = 1.695 moles of Ostart subscript, 2, end subscript.

**Step 4: Convert moles back to ‘grams’ using molar mass of Ostart subscript, 2, end subscript**

Molar mass of Ostart subscript, 2, end subscript = 2 (atomic mass of O) = 2 (15.999) = 31.998 g
i.e. 1 mole of Ostart subscript, 2, end subscript = 31.998 g of Ostart subscript, 2, end subscript
therefore, 1.695 moles of Ostart subscript, 2, end subscript = 31.998 g/mol x 1.695 mol = 54.236 g

*Answer: If 138.6 g of KClOstart subscript, 3, end subscript is heated and decomposes completely, 54.236 g of oxygen gas will be produced*

### Zadanie 2

How many grams of iodine must react to give 4.63 grams of ferric iodide?

**Step 1: Ensure that the reaction is balanced**

The above reaction is balanced!!!

**Step 2: Convert grams of FeIstart subscript, 3, end subscript to moles**

For this we first need to calculate the molar mass of FeIstart subscript, 3, end subscript = 1 (atomic mass of Fe) + 3 (atomic mass of I) = 1 (55.845) + 3 (126.904) = 55.845 + 380.712 = 436.557 g

i.e. 436.557 g of FeIstart subscript, 3, end subscript = 1 mole of FeIstart subscript, 3, end subscript

therefore, 1 g of FeIstart subscript, 3, end subscript = (1 mole/ 436.557 g) x 1 g

so, 4.63 g of FeIstart subscript, 3, end subscript = (1 mole/ 436.557 g) x 4.63 g = 0.0106 moles

**Step 3: Use mole ratios (from the balanced chemical reaction) to calculate the moles of the relevant reactant or product**

In this example, we need to calculate the mass of Istart subscript, 2, end subscript that must react. So we first need to calculate the moles of Istart subscript, 2, end subscript that would be required to produce 0.0106 moles of FeIstart subscript, 3, end subscript

Looking at the balanced chemical reaction, we can infer that 3 moles of Istart subscript, 2, end subscript are required to produce 2 moles of FeIstart subscript, 3, end subscript

So, 0.0106 moles of FeIstart subscript, 3, end subscript will require [(3 moles of Istart subscript, 2, end subscript / 2 moles of FeIstart subscript, 3, end subscript) x 0.0106 moles of FeIstart subscript, 3, end subscript] = 0.0159 moles of Istart subscript, 2, end subscript

**Step 4: Convert moles back to ‘grams’ using molar mass of Istart subscript, 2, end subscript**

Molar mass of Istart subscript, 2, end subscript = 2 (atomic mass of I) = 2 (126.904) = 253.808 g

i.e. 1 mole of Istart subscript, 2, end subscript = 253.808 g of Istart subscript, 2, end subscript

therefore, 0.0159 moles of Istart subscript, 2, end subscript = (253.808 x 0.0159) g = 4.035 g

**Answer: 4.035 grams of iodine must react to yield 4.63 grams of ferric iodide.**

### Zadanie 3

How many grams of Hstart subscript, 2, end subscriptO will be produced when you burn 25 grams of methane?

**Step 1: Ensure that the reaction is balanced**

In this case the reaction is not balanced. After balancing, the equation looks like

**Step 2: Convert grams of CHstart subscript, 4, end subscript to moles**

For this we first need to calculate the molar mass of CHstart subscript, 4, end subscript = 1 (atomic mass of C) + 4 (atomic mass of H) = 1 (12.011) + 4 (1.008) = 12.011 + 4.032 = 16.043 g

i.e. 16.043 g of CHstart subscript, 4, end subscript = 1 mole of CHstart subscript, 4, end subscript

therefore, 1 g of CHstart subscript, 4, end subscript = (1 mole/ 16.043 g) x 1 g

so, 25 g of CHstart subscript, 4, end subscript = (1 mole/ 16.043 g) x 25 g = 1.558 moles

**Step 3: Use mole ratios (from balanced chemical reaction) to calculate the moles of the relevant reactant or product**

In this example, we need to calculate the mass of Hstart subscript, 2, end subscriptO produced. So we first need to calculate the moles of Hstart subscript, 2, end subscriptO that would be produced from 1.558 moles of CHstart subscript, 4, end subscript

Looking at the

*balanced*chemical reaction, we can infer that 1 mole of CHstart subscript, 4, end subscript produces 2 moles of Hstart subscript, 2, end subscriptOSo, 1.558 moles of CHstart subscript, 4, end subscript will produce [(2 moles of Hstart subscript, 2, end subscriptO / 1 mole of CHstart subscript, 4, end subscript ) x 1.558 moles of CHstart subscript, 4, end subscript] = 3.116 moles of Hstart subscript, 2, end subscriptO

**Step 4: Convert moles back to ‘grams’ using molar mass of Hstart subscript, 2, end subscriptO**

Molar mass of Hstart subscript, 2, end subscriptO = 2 (atomic mass of H) + 1 (atomic mass of O) = 2 (1.008) + 1 (15.999) = 18.015 g

i.e. 1 mole of Hstart subscript, 2, end subscriptO = 18.015 g of Hstart subscript, 2, end subscriptO

therefore, 3.116 moles of Hstart subscript, 2, end subscriptO = (18.015 x 3.116) g = 56.134 g

**Answer: 56.134 g of Hstart subscript, 2, end subscriptO will be produced when we burn 25 grams of methane**

After going through the above problems, we should now have a fair idea of how to calculate the amount of a reactant used or the amount of a product formed in a chemical reaction from a given amount of any one of the reactants or products.

## Concept of ‘limiting reactant’ and ‘excess reactant’

**Limiting Reactant**is defined as the reactant in a chemical reaction that limits the amount of product that is formed. The reaction will stop when all of the limiting reactant is consumed.

**Excess Reactant**is defined as the reactant in a chemical reaction that is in excess and remains unconsumed when a reaction stops because the limiting reactant has been completely consumed.

Let’s understand this concept with an example. Suppose there are 4 kids and 9 pairs of flip-flops. Each kid will wear a pair of flip-flops, leaving behind 5 pairs of flip-flops.

No matter how many pairs of flip-flops there are, only 4 pairs of flip-flops can be worn because there are only 4 kids. Likewise in chemical reaction, when the limiting reagent (kid) is used up, the reaction will stop and the unconsumed excess reagent (pairs of flip-flops) will remain in the reaction mixture.

*Now the key task is to be able to identify which reactant is the limiting reactant in a given chemical reaction.*

*The easiest way to identify which reactant is the ‘limiting reactant’ and which one is the ‘excess reagent’ is by taking the number of moles of each reactant and dividing it by its molar coefficient, as indicated in the balanced chemical reaction. The substance that has the lowest value is the limiting reagent.*

As illustrated above, in an ideal world 36 popcorns (product) should pop out of 36 seeds (reactant). In this case, ‘36’ is referred to as the ‘

*theoretical yield*’. But actually, only 27 popcorns pop. Here, ‘27’ is referred to as the ‘actual yield. The percent yield of this popcorn making process would be defined as ( start fraction, a, c, t, u, a, l, y, i, e, l, d, divided by, t, h, e, o, r, e, t, i, c, a, l, y, i, e, l, d, end fraction x 100) = 27/ 36 x 100 = 75%. In other words, the popcorn making process was only 75% efficient.The above analogy can be applied to any chemical reaction. As we have learned in the first part of this tutorial, the amount of product formed can be theoretically calculated from the given amount of the limiting reagent and the mole ratios of the reactants and products. This is called the

*theoretical yield*of a reaction.However, the amount of product actually formed in a reaction is usually less than the theoretical yield and is referred to as the actual yield. This is because reactions often have "side reactions" that also compete for reactants and produce undesired products.

*To evaluate the efficiency of the reaction, chemists compare the actual and theoretical yields by calculating the percent yield of a reaction:*Percent yield of a reaction = start fraction, a, c, t, u, a, l, y, i, e, l, d, divided by, t, h, e, o, r, e, t, i, c, a, l, y, i, e, l, d, end fraction x 100

Let’s attempt a problem now.

### Zadanie 4

Let’s take the example of preparation of nitrobenzene (Cstart subscript, 6, end subscriptHstart subscript, 5, end subscriptNOstart subscript, 2, end subscript), starting with 35.15 g of benzene (Cstart subscript, 6, end subscriptHstart subscript, 6, end subscript) and an excess of nitric acid (HNOstart subscript, 3, end subscript). The balanced chemical reaction for this is

Calculate the percent yield if your preparation yielded 52.145g of nitrobenzene.

Let’s first calculate the theoretical yield of Cstart subscript, 6, end subscriptHstart subscript, 5, end subscriptNOstart subscript, 2, end subscript

**Krok 1: Convert grams of benzene to moles**

Molar mass of Cstart subscript, 6, end subscriptHstart subscript, 6, end subscript = 6 (atomic mass of C) + 6 (atomic mass of H) = 6 (12.011) + 6 (1.008) = 72.066 + 6.048 = 78.114 g

i.e. 78.114 g of Cstart subscript, 6, end subscriptHstart subscript, 6, end subscript = 1 mole of Cstart subscript, 6, end subscriptHstart subscript, 6, end subscript

therefore, 1 g of Cstart subscript, 6, end subscriptHstart subscript, 6, end subscript = (start fraction, start text, space, m, o, l, e, end text, divided by, 78, point, 114, start text, space, g, end text, end fraction) x 1 g

so, 35.15 g of Cstart subscript, 6, end subscriptHstart subscript, 6, end subscript = (start fraction, 1, start text, space, m, o, l, e, end text, divided by, 78, point, 114, start text, space, g, end text, end fraction) x 35.15 g = 0.449 moles

**Krok 2: Determine the theoretical yield of nitrobenzene (convert moles of benzene to moles of nitrobenzene)**

Looking at the

*balanced*chemical reaction, we can infer that 1 mole of Cstart subscript, 6, end subscriptHstart subscript, 6, end subscript produces 1 mole Cstart subscript, 6, end subscriptHstart subscript, 5, end subscriptNOstart subscript, 2, end subscript.So, 0.449 moles of Cstart subscript, 6, end subscriptHstart subscript, 6, end subscript will

*theoretically*produce start fraction, 1, start text, space, m, o, l, e, space, o, f, space, end text, C, start subscript, 6, end subscript, H, start subscript, 5, end subscript, N, O, start subscript, 2, end subscript, divided by, 1, start text, space, m, o, l, e, space, o, f, space, end text, C, start subscript, 6, end subscript, H, start subscript, 6, end subscript, end fraction = 0.449 moles of Cstart subscript, 6, end subscriptHstart subscript, 5, end subscriptNOstart subscript, 2, end subscript**Krok 3: Convert moles back to grams using molar mass of Cstart subscript, 6, end subscriptHstart subscript, 5, end subscriptNOstart subscript, 2, end subscript**

Molar mass of Cstart subscript, 6, end subscriptHstart subscript, 5, end subscriptNOstart subscript, 2, end subscript= 6 (atomic mass of C) + 5 (atomic mass of H) + 1 (atomic mass of N) + 2 (atomic mass of O) = 6 (12.011) + 5 (1.008) + 1 (14.0067) + 2 (15.999) = 72.066 + 5.04 + 14.0067 + 31.998 = 123.111g

i.e. 1 mole of Cstart subscript, 6, end subscriptHstart subscript, 5, end subscriptNOstart subscript, 2, end subscript = 123.111 grams of Cstart subscript, 6, end subscriptHstart subscript, 5, end subscriptNOstart subscript, 2, end subscript

Therefore, 0.449 moles of Cstart subscript, 6, end subscriptHstart subscript, 5, end subscriptNOstart subscript, 2, end subscript = (123.111 x 0.449) g = 55.277g

**Krok 4: Determine the percent yield.**

Percent yield of a reaction = actual yield/theoretical yield x 100%

**Answer:**Percent yield of nitrobenzene = (52.145g/55.277g) x 100% = 94.334%

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