The Stoichiometry Calculator is a free online tool that displays a balanced equation for a chemical equation. STUDYQUERIES’s online stoichiometry calculator tool makes calculations faster, and it displays the balanced equation in a fraction of a second.
How to Use Stoichiometry Calculator?
Here is how to use the Stoichiometry calculator:
- Step 1: Enter the chemical equation into the input field
- Step 2: Click the “Submit” button to get the results
- Step 3: The balanced chemical equation will be displayed in the new window
Stoichiometry Calculator
What is a Stoichiometry?
Stoichiometry is a section of chemistry that uses relationships between reactants and/or products of a chemical reaction to determine desired quantitative data. Stoichiometry literally translates as the measurement of elements, since stoikhein means element and metron means to measure.
For stoichiometry to be useful in running calculations about chemical reactions, a basic understanding of the relationships between products and reactants, as well as their reasons, is needed, which entails learning how reactions are balanced.
What is a Chemical Equation?
To represent the various chemicals in chemistry, we use symbols. It is important to become familiar with these basic symbols in order to be successful in chemistry. As an example, the symbol “C” represents an atom of carbon, and “H” represents an atom of hydrogen.
We would use the notation “NaCl” to represent a molecule of table salt, sodium chloride, since “Na” stands for sodium and “Cl” stands for chlorine. In this case, chlorine is called “chloride” because of its connection to sodium. Nomenclature, or naming schemes, should have been covered in earlier readings.
Chemical equations are expressions of chemical processes. They are, for example:
$$AgNO_3(aq) + NaCl(aq)\longrightarrow AgCl(s) + NaNO_3(aq)$$
AgNO3 and NaCl are mixed in this equation. According to the equation, the reactants (AgNO3 and NaCl) react through some process to form products (AgCl and NaNO3). By undergoing a chemical reaction, they are fundamentally changed.
There are often chemical equations that show what state each substance is in. The (s) sign indicates that the compound is solid. The (l) sign indicates that the substance is liquid. The (aq) sign stands for aqueous in water and means the compound is dissolved in water. Lastly, the (g) sign indicates that the compound is a gas.
In all chemical equations, coefficients are used to indicate the relative amounts of each substance. An amount can represent either the relative number of molecules or the relative number of moles (described below). One (1) is assumed in the absence of a coefficient.
It is not uncommon for a variety of information to be written above or below the arrows. By providing information such as a temperature value, we can determine what conditions are needed for a reaction to take place. The notation above and below the arrows in the graphic below shows that this reaction must occur at a chemical Fe2O3, a temperature of 1000°C, and a pressure of 500 atmospheres.
The following graphic illustrates many of the concepts described above:
The Mole
From the equation above, we can determine the number of moles of reactants and products. A mole simply represents Avogadro’s number (6.022 x 1023) of molecules. A mole is similar to the term dozen. When you have a dozen carrots, you have twelve of them.
Likewise, if you have a mole of carrots, you have 6.022 x 1023 carrots. Since there are no numbers in front of the terms in the equation above, each coefficient is assumed to be one (1). Therefore, you have the same number of moles of AgNO3, NaCl, AgCl, NaNO3.
It is often necessary to convert between moles and grams of a substance. The conversion can be done easily when the atomic and/or molecular mass of the substance(s) is known. Based on the atomic or molecular mass of a substance, a mole of that substance is equal to a gram of the substance.
Calcium, for example, has an atomic mass of 40. In other words, 40 grams of calcium equals one mole, 80 grams equals two moles, etc.
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Balancing Chemical Equations
There are times, however, when we have to do some work before we can use the coefficients of the terms to determine the relative number of molecules of each compound. Such is the case when the equations are not balanced properly. Let’s look at the following equation:
$$Al + Fe_3O_4\longrightarrow Al_2O_3 + Fe$$
Because no coefficients are mentioned for any of the terms, we can assume that one (1) mole of Al reacts with one (1) mole of Fe3O4 to produce one (1) mole of Al2O3. Assuming this is true, the reaction would be quite spectacular: an aluminum atom would appear out of nowhere, and two (2) iron atoms and one (1) oxygen atom would magically disappear.
We need to make sure that the number of atoms of each element in the reactants equals the number of atoms in the products. To accomplish this, we have to find the relative number of molecules of each term as a function of its coefficient.
Trial and error is the most effective way to balance a simple chemical equation. The number of atoms in an equation can be calculated in a number of different ways, but no matter how you do it, you need to know how to count them in. Let’s examine the following example.
$$2Fe_3O_4$$
The term refers to two (2) molecules of Fe3O4. Three (3) Fe atoms are found in each molecule of this substance. Therefore, in every two (2) molecules of the substance, there must be six (6) Fe atoms. Additionally, one (1) molecule of the substance contains four (4) oxygen atoms, so there must be eight (8) oxygen atoms in two (2) molecules.
Let’s try to balance the equation we discussed earlier:
$$Al + Fe_3O_4\longrightarrow Al_2O_3 + Fe$$
Developing a strategy can be challenging, but here is one way to approach a situation like this. Take note of the number of atoms on each side of the reaction.
The first step to balancing is to determine a term. Taking a closer look at this problem, it appears that oxygen will be the most difficult element to balance, so we’ll start with oxygen. To balance the oxygen terms, simply follow these steps:
$$Al + 3Fe_3O_4\longrightarrow 4Al_2O_3 + Fe$$
Note that the subscript times the coefficient will give the number of atoms in that element. The reactant has a coefficient of three (3) multiplied by a subscript of four (4), giving 12 oxygen atoms. The product side has a coefficient of four (4) multiplied by a subscript of three (3), giving 12 oxygen atoms. At this point, the oxygens have been balanced.
Balance the equation with another term. Let’s choose iron, Fe. We add a nine (9) coefficient in front of the Fe since there are nine (9) iron atoms in the term in which the oxygen is balanced. Therefore, we have:
$$Al + 3Fe_3O_4\longrightarrow 4Al_2O_3 + 9Fe$$
Balancing the last term. On the reactant side, since we had eight (8) aluminum atoms on the product side, we need another eight (8) in front of the Al term.
Our equation is now balanced, and the answer is:
$$8Al + 3Fe_3O_4\longrightarrow 4Al_2O_3 + 9Fe$$
Limiting Reagents
The reactant that runs out first in a reaction between two or more substances sometimes occurs before the other. This is known as the “limiting reagent”. It is often necessary to identify the limiting reagent in a problem.
Example: The chemist has 6.0 grams of C2H2 and an unlimited supply of oxygen, and he wishes to produce as much CO2 as he can. How much oxygen should she add to the reaction if she uses the equation below?
$$2C_2H_2(g) + 5O_2(g)\longrightarrow 4CO_2(g) + 2 H_2O(l)$$
This problem can be solved by determining how much oxygen should be added if all of the reactants have been used up (this is the best way to get the maximum amount of CO2).
We first calculate the number of moles of C2H2 in 6.0 g of C2H2. We need to look at the periodic table in order to calculate the moles and know that 1 mole of C weighs 12.0 g and 1 mole of H weighs 1.0 g. Thus, 1 mole of C2H2 weighs 26 grams (2 × 12 grams + 2 × 1 gram).
$$6.0 g\ C_2H_2\times \frac {1\ mol\ C_2H_2}{(24.0 + 2.0)g\ C_2H_2}=0.25\ mol\ C_2H_2$$
In order to find the total molecules of oxygen, we must multiply the result by 5/2, since there are five (5) molecules of oxygen to every two (2) molecules of C2H2. Once we have converted to grams, we can determine how much oxygen needs to be added:
$$0.25 g\ C_2H_2\times \frac {5\ mol\ O_2}{2\ mol\ C_2H_2}=20g\ O_2$$
Percent Composition
Using the percent by mass of products or reactants in a chemical equation, it is possible to calculate the mole ratios (also known as mole fractions) between the terms.
$$percentage\ by\ mass = \frac {mass\ of\ part}{mass\ of\ the\ whole}$$
There are two types of percent composition problems, problems in which you are given the formula (or the weight of each part) and asked to calculate the percentage of each element, and problems in which you are given the percentages and asked to calculate the formula.
There are many possible solutions to percent composition problems. Doubling the answer is always an option. The proportions of CH and C2H2 are the same, but they are different compounds. Normally, compounds are given in their simplest form, where the ratio between the elements is as close to the empirical formula as possible.
You can convert percentages to grams for calculating the empirical formula based on percent composition. It is usually easier to assume you have 100 grams so 54.3% would become 54.3 grams. Having converted the masses to moles, we can calculate mole ratios. The ratios need to be reduced to whole numbers.
Dividing all the terms by the smallest number of moles is a good technique. To write the empirical formula, the ratio of the moles is transferred.
Example: Given a compound containing 47.3% C (carbon), 10.6% H (hydrogen), and 42.0% S (sulfur), what is its empirical formula?
We must convert all of our percents to masses in order to solve this problem. Let’s assume we have 100 grams of this substance. We convert to moles as follows:
$$Carbon: \frac{47.3\ grams}{1}\times \frac{1\ mole}{12.01\ grams}= 3.94\ moles$$
$$Hydrogen: \frac{10.6\ grams}{1}\times \frac{1\ mole}{1.008\ grams}= 10.52\ moles$$
$$Sulfur: \frac{42.0\ grams}{1}\times \frac{1\ mole}{32.07\ grams}= 1.310\ moles$$
Then we divide by the number of moles of sulfur, since that is the smallest number, to get an even ratio between the elements:
$$Carbon: \frac{3.94}{1.310} = 3$$
$$Hydrogen: \frac{10.52}{1.310} = 8$$
$$Sulfur: \frac{1.310}{1.310} = 1$$
So we have: $$C_3H_8S$$
Example: Calculate the percentage of hydrogen sulfate, H2SO4, by mass.
We first need to calculate the total mass of the compound by looking at the periodic table. That gives us:
$$[2\times (1.008) + 32.07 + 4\times (16.00)] \frac{g}{mol} = 98.09 \frac{g}{mol}$$
The weight fraction of each element over the total mass (which we just found) must be multiplied by 100 to get a percentage.
$$Hydrogen: \frac{2\times (1.008)}{98.09} = {2.016}{98.09} = 0.0206 \times 100 = 2.06\ Percent$$
$$Sulfur: \frac{32.07}{98.09}= 0.327\times 100 = 32.7\ Percent$$
$$Oxygen: \frac {4\times (16.00)}{98.09}= \frac{64.00}{98.09}= 0.652 \times 100 = 65.2\ Percent$$
We can now verify that the percentages add up to 100%
$$65.2 + 2.06 + 32.7 = 99.96$$
It is essentially 100, so we know everything has worked, and we probably did not make any careless mistakes. Therefore, H2SO4 is made up of 2.06% H, 32.7% S, and 65.2% O by mass.
Empirical Formula and Molecular Formula
An empirical formula is the simplest form of a compound, whereas a molecular formula represents the term in a chemical equation. Either the empirical formula and molecular formula are the same, or the molecular formula is any positive integer multiple of the empirical formula. An example of an empirical formula is AgBr, Na2S, C6H10O5. The following are examples of molecular formulas: P2, C2O4, C6H14S2, H2, C3H9.
From the mass or percentage composition of any compound, one can calculate the empirical formula. Percent composition has already been discussed above. The only thing we are doing is eliminating the step of converting percentage to mass.
Example: Calculate the empirical formula for a compound containing 43.7 grams of phosphorus and 56.3 grams of oxygen.
Our first step is to convert to moles:
$$\frac {43.7\ grams\ P}{1}\times \frac{1\ mol}{30.97\ grams}= 1.41\ moles$$
$$\frac {56.3\ grams\ O}{1}\times \frac{1\ mol}{16.00\ grams} = 3.52\ moles$$
Next, we divide the moles to try to get an even ratio.
$$Phosphorus: \frac {1.41}{1.41}= 1.00$$
$$Oxygen: \frac {3.52}{1.41}= 2.50$$
When we divide, we did not get whole numbers so we must multiply by two (2). The answer = $$P_2O_5$$
When we have the empirical formula, calculating the molecular formula is easy. We need only divide the molecular mass of a compound by its empirical formula if we know its empirical formula. The same can be done with one of the elements in the formula; simply divide the mass of that element in one mole of the compound by its mass in the empirical formula. If the result is a natural number, the mass ratio is correct.
Density
The density of a substance is its mass per unit volume. The term is commonly used in chemistry.
Concentrations of Solutions
Concentration determines the “strength” of a solution. Typically, a solution is the dissolution of something solid in a liquid, such as salt dissolving in water. Often, it is also necessary to figure out how much water to add to a solution to change its concentration.
Molarity
Molarity is typically used to describe the concentration of a solution. Molecularity is defined as the number of moles of solute (what is actually dissolved in the solution) divided by the volume in liters of the solution (which contains both the dissolved substance and the solution itself).
$$Molarity = \frac {moles\ of\ solute}{Volume\ in\ liters\ of\ solution}$$
The molarity term is probably most commonly used since measuring the volume of liquid is relatively easy.
Example: If 5.00 g of NaOH are dissolved in 5000 mL of water, what is the molarity of the solution?
One of our first steps is to convert the amount of NaOH given in grams into moles:
$$= \frac {5.00g\ NaOH}{1}\times \frac {1\ mole}{(22.9 + 16.00 + 1.008)g}= 0.125\ moles$$
Now we simply use the definition of molarity: moles/liters to get the answer
$$Molarity =\frac {0.125\ moles}{5.00\ L\ of\ sol^n}= 0.025\frac {mol}{L}$$
So the molarity (M) of the solution is $$0.025\frac {mol}{L}.$$
Molality
Molality is another way to measure concentration. Molality is calculated as moles of solute divided by kilograms of solvent (the substance in which the solute is dissolved).
$$Molality =\frac {moles\ of\ solute}{Weight\ in\ kg\ of\ solvent}$$
When extreme temperatures are present, molality may be substituted for molarity because the volume can shrink or expand.
Example: If the molality of a solution of C2H5OH in water is 1.5 and the mass of the water is 11.7 kg, calculate how much C2H5OH has been added to the solution in grams.
First, we need to substitute what we know into the equation. Our next step is to solve for what we don’t know: moles of solute. We can convert moles to grams by looking at the periodic table after we know the moles of the solute.
$$Molality =\frac {moles\ of\ solute}{Weight\ in\ kg\ of\ solvent}$$
We can now calculate the answer using the definition of molarity: moles/liters
$$Molality =\frac {moles\ of\ solute}{Weight\ in\ kg\ of\ solvent}$$
$$1.5\frac {moles}{Kg} = \frac {moles\ of\ solute}{11.7 Kg}$$
$$1.5\frac {moles}{Kg}\times 11.7 Kg = 17.55\ Moles$$
$$\frac {17.55\ Moles}{1}\times \frac {(2\times 12.01) + (6\times 1.008) + 16}{1\ Mol}= 808.5 g\ C_2H_5OH$$
Converting between molarity and molality is possible. Density is the only information required.
Example: If the molarity of a solution is 0.30 M, calculate the molality of the solution knowing that the density is 3.25 g/mL.
To make the calculations easier, we can assume one (1) liter of the solution. We need to get from the molarity units of mol/L to the molality units of mol/kg. Keeping in mind that a liter is 1000 mL and a kilogram is 1000 grams, let’s work the problem this way. There will only be an accurate conversion at small molarities and molalities.
$$\frac{0.3 mol}{1\ L}\times \frac{1 mL}{3.25 g}\times \frac {1 L}{1000 mL}\times \frac{1000 g}{1 kg}= 0.09 \frac{Moles }{kg}$$
Using molality, it is also possible to calculate colligative properties, such as boiling point depression. The equation for temperature depression is as follows.
$$\Delta T= K_f\times m$$
Where:
- ΔT is temperature depression (for freezing point) or temperature expansion (for boiling point) (°C)
- Kf is the freezing point constant (kg °C/mol)
- m is molality in mol/kg
As an example: What is the molality of the solution if the freezing point of the saltwater used for roads is -5.2° C? (The Kf for water is 1.86 °C/m.)
For this simple problem, we are just plugging numbers into an equation. The one thing we do need to know is that water usually freezes at 0° C.
$$\Delta T= K_f\times m$$
$$\frac {ΔT}{K_f} = m$$
$$m = \frac {5.2}{1.86}$$
$$m = 2.8 \frac {moles}{kg}$$
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Conclusion
A balanced chemical equation shows us the numerical relationships between each species involved in the chemical reaction. Using these mathematical relationships, we can calculate mole ratios, allowing us to convert between the amounts of reactants and/or products (and thus solve stoichiometry problems!).
FAQs
How do you calculate stoichiometry?
Almost all stoichiometric problems can be solved by following four simple steps:
- You need to balance the equation.
- Convert units of a given substance into moles.
- By using the mole ratio, calculate the moles of substance yielded by the reaction.
- Convert moles of desired substance to desired units.
What is stoichiometry in chemistry?
Stoichiometry is exactly what it sounds like. A chemical reaction is measured by the number of moles (and therefore mass) of various products and reactants. Reactions involving chemical compounds must be balanced, or in other words, have the same number of atoms in the products as in the reactants.
Why is stoichiometry so hard?
The concept of stoichiometry can be challenging because it involves a number of individual skills. You must master problem-solving skills and plan a problem-solving strategy if you want to be successful. Before going on to Calculating Molar Mass, master each of these skills.
What is the mole ratio?
Among the most common types of stoichiometric relationships is the mole ratio, which describes the amount of each substance in a reaction in moles. The mole ratio of a pair of substances can be determined by examining the coefficients in front of each species in a balanced chemical equation.
What is the first step in all stoichiometry calculations?
It is essential in every stoichiometric problem to ensure that the chemical reaction you are dealing with is balanced and that you understand the concept of a ‘mole’ and the relationship between ‘amount (grams)’ and ‘moles’.