Conservation of mass

In a nutshell

When writing chemical equations, there is always the same number of atoms on the right-hand side as there is on the left-hand side. This is because no atoms are lost or made during a reaction due to the law of conservation of mass.

Equations

Chemical equations

In a chemical reaction, bonds between atoms in the reactants are broken and new bonds are made between atoms to form the products. The number of atoms stays the same and therefore the sum of all the relative formula masses stays the same on both sides.

The relative formula mass of a compound is the sum of relative atomic masses:

​$$relative ~formula ~mass = \sum relative ~atomic ~masses \\ \ \\ Mr=\sum Ar$$

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Note: $$Ar$$ of atoms can be found by looking at the biggest number of each square in the periodic table and the relative formula mass is sometimes called the relative molecular mass.​​

PROCEDURE

Tip: The molar ratio is the number in front of the compound. If there is no number in front, the molar ratio is one.

Example

$$magnesium+ hydrochloric~acid \rightarrow magnesium ~chloride + hydrogen \\ \ \\ Mg(s)+ 2HCl(aq)\rightarrow MgCl_2(aq) + H_2(g)$$

On the left-hand side (reactants):  

On the right-hand side (products): 

​ $$\sum({Mr_{reactant} \times molar~ ratio}) =\sum({Mr_{product} \times molar~ ratio})=\underline{97}$$​ due to the law of conservation of mass.  

Calculations

The law of conservation states that no atoms are made or destroyed during a reaction; it can be used to calculate the unknown mass of a product.

PROCEDURE 

Example

In the following reaction, product $$D$$ escapes from the reaction vessel and therefore its mass cannot be measured by a measuring balance. However, its mass can be calculated instead, using the information below. Calculate the mass of $$D$$​.

​$$A(s)+B(s)\rightarrow C(s)+D(g)+E(s)\newline mass~of~A= 9~g\newline mass~of~B= 6~g \newline mass~of~C= 5~g \newline mass~of~E=2~g$$

Calculate the the sum of reactant masses $$(A~and~B)$$​:

​$$mass~of~A+mass~of~B= 9+6 =15~g$$

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Add all the known masses of products $$(C~and~E)$$:

$$mass~of~C+mass~of~E=5+2=7~g$$

Subtract the known masses of products from the sum of reactant masses to get the mass of the unknown product $$(D)$$​:

​$$mass~of~D = 15-7=8~g$$​​

The mass of the gaseous product D is $$\underline{8 \, g}$$.

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State symbols

Identifying state symbols in equations is important for understanding why it may seem like mass has been lost/gained.

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Closed systems

In a closed system, the reactants and products cannot escape from the reaction vessel. The mass of substances in the reaction vessel therefore remains constant. This is due to either the equipment that is being used or the nature of the reaction.

Example

​$$hydrochloric~acid+ sodium ~hydroxide\rightarrow sodium~chloride+ water\\ \ \\HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+ H_2O(l)$$

In this reaction, all the reactants and products remain in aqueous solution $$(aq)$$ so they do not escape from the beaker.

Tip: Aqueous solution means dissolved in water.

Non-enclosed systems

In a non-enclosed system, the reactants and products can escape from the reaction vessel. The mass of substances in the reaction vessel can therefore change. Gas reactants and gas products in open non-enclosed systems can make it seem like mass has been lost/gained throughout a reaction.

Tip: Non-enclosed systems are sometimes called open systems.

Gas reactants  

The presence of gas reactants in a non-enclosed system can make it seem like mass has been gained.

PROCEDURE

Example

​$$magnesium + oxygen \rightarrow magnesium ~oxide~ \\ \ \\ 2Mg(s) + O_2(g)\rightarrow 2MgO(s)$$

In this example, the oxygen gas at the start is not measurable by a balance so when it gets incorporated into the solid magnesium oxide product, which is measurable by a balance, it makes it look like the mass has increased.

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Gas products

The presence of gas products in a non-enclosed system can make it seem like mass has been lost because the gas escapes from the reaction vessel.

PROCEDURE

Example

​$$calcium ~carbonate\rightarrow calcium~oxide +carbon~dioxide\\ \ \\CaCO_3(s)\rightarrow CaO(s )+CO_2(g)$$​​

In this example, there are no gas reactants but there is a gas product, carbon dioxide, which escapes from the reaction and causes the mass in the reaction vessel to decrease. 

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