Topic: Calculation of molecular masses - ratio of the number of atoms, mass ratio

Target group

Elementary school student (grades 7. and 8.)

Core curriculum:

Elementary school. Chemistry.

III. Chemical reactions. Pupil:

6. calculates molecular masses of elements occurring in the form of molecules and chemical compounds.

General aim of education

The student calculates the mass ratio of individual elements in a chemical compound.

Key competences

• communication in the mother tongue;

• communication in foreign languages;

• mathematical competence and basic competences in science and technology;

• digital competence;

• learning to learn.

Criteria for success
The student will learn:

• to calculate the ratio of the number of atoms of individual elements in a chemical compound;

• to calculate the atomic mass of the X element, knowing the molecular weight of the element X oxide with the given sum formula;

• to determine the mass ratios of elements in chemical compounds.

Methods/techniques

• expository

  • talk.

• activating

  • discussion.

• exposing

  • film.

• programmed

  • with computer;

  • with e‑textbook.

• practical

  • exercices concerned.

Forms of work

• individual activity;

• activity in pairs;

• activity in groups;

• collective activity.

Teaching aids

• e‑textbook;

• notebook and crayons/felt‑tip pens;

• interactive whiteboard, tablets/computers.

Lesson plan overview

Introduction

1. The teacher hands out Methodology Guide or green, yellow and red sheets of paper to the students to be used during the work based on a traffic light technique. He presents the aims of the lesson in the student's language on a multimedia presentation and discusses the criteria of success (aims of the lesson and success criteria can be send to students via e‑mail or posted on Facebook, so that students will be able to manage their portfolio).

2. The teacher together with the students determines the topic – based on the previously presented lesson aims – and then writes it on the interactive whiteboard/blackboard. Students write the topic in the notebook.

Realization

1. The teacher introduces students to the issues implemented during the lesson. It presents an illustration of the ratio of the number of hydrogen atoms, phosphorus and oxygen in phosphoric acid from abstract. He explains how the ratio of the number of atoms of particular elements in a chemical compound is calculated.

2. Students, working in pairs, analyze the data in the table „Ratio of the number of atoms (or ions) of particular elements in selected chemical compounds”. The indicated couples discuss their observations on the class forum.

3. The teacher gives several selected compounds, e.g. CIndeks dolny 6₆HIndeks dolny 12₁₂OIndeks dolny 6₆, NH Indeks dolny 2₂-CHIndeks dolny 2₂-COOH, NHIndeks dolny 3₃, Pb(OH)Indeks dolny 2₂, CIndeks dolny 6₆HIndeks dolny 12₁₂, CIndeks dolny 4₄HIndeks dolny 10₁₀, and asks willing students to calculate the ratio of the number of individual atoms elements on the blackboard.

4. The teacher plays the presentation titled „Calculating the mass ratio of individual elements in a chemical compound”. Students note calculations made on the example of calcium oxide (CaO), phosphorus oxide (PIndeks dolny 4₄OIndeks dolny 10₁₀), ethane (CIndeks dolny 2₂HIndeks dolny 6₆), pentyne (CIndeks dolny 5₅HIndeks dolny 8₈).

5. Students read fragment titled „How we calculate the mass ratio of individual elements in a chemical compound?” and answer the question in the title of the fragment. Together with the teacher, they discuss the illustration „The mass ratio of elements in water” and the table „Mass ratio of elements in selected chemical compounds”.

6. The teacher gives examples of chemical compounds: HIndeks dolny 2₂SOIndeks dolny 3₃, HIndeks dolny 2₂S, HIndeks dolny 2₂COIndeks dolny 3₃, HNOIndeks dolny 3₃, Fe(OH)Indeks dolny 3₃, Cu(OH)Indeks dolny 2₂, NHIndeks dolny 3₃. Volunteers calculate the mass ratio of individual elements in a chemical compound.

7. The lecturer displays the presentation „Determining the chemical formula based on the mass ratio of elements in this compound” and asks students to pay attention to the oxide in which the mass ratio of nitrogen to oxygen is 7:12.

8. Students carry out exercise number 1 and exercise number 2. The teacher discusses the correct solutions with them.

Summary

1. The teacher chooses one student by random method and asks him or her to explain in own words the meaning of a given word or concept learned during the lesson.

2. The teacher asks a willing student to summarize the lesson from his point of view. He asks other students if they would like to add anything to their colleague's statements.

Homework

1. Using the generator in the abstract, create a multiple‑choice test based on today's lesson.

The following terms and recordings will be used during this lesson

Terms

Texts and recordings

Calculation of molecular masses - ratio of the number of atoms, mass ratio

Based on the formula of a chemical compound, the ratio of the number of atoms (or ions) of particular elements forming this compound can be determined. For example in hydrogen chloride with the formula: $\text{HCl}$ for one hydrogen atom there is one chlorine atom. The ratio of the number of hydrogen atoms to the number of chlorine atoms is 1: 1. In the case of water, the total formula is as follows:${\text{H}}_{\text{2}}\text{O}$, the ratio of the number of hydrogen atoms to the number of oxygen atoms is 2: 1.

The summary formula of a chemical compound contains information about which elements are included in its composition and in what quantitative relations they combine with each other. With this knowledge, it is easy to calculate mass ratio individual elements. This value is defined as the ratio of the mass of atoms of individual elements included in the chemical compound. In the hydrogen chloride molecule of the formula $\text{HCl}$ there is one atom of hydrogen with an atomic mass of 1 u and one atom of chlorine with an atomic mass of 35.5 u. The mass ratio of hydrogen to chlorine in the hydrogen chloride molecule is therefore $1:35,5$.

In a compound of the general formula:

${\text{A}}_{\text{x}}{\text{B}}_{\text{y}}$

the mass ratio is as follows:

$\mathrm{x}·\text{ atomic weight of element }\mathrm{A}:\mathrm{y}·\text{ atomic weight of element }\mathrm{B}$

• The molecular weight of a chemical compound is equal to the sum of the mass of the atoms of the elements that make up the molecule (if the compound is covalent) or the smallest set of repeating ions (if the compound has an ionic structure). It is expressed in atomic mass units.

• The ratio of the masses of individual elements that make up a chemical compound is called the mass ratio of the elements in a compound.