Question

A binary compound between an unknown element \(\mathrm{E}\) and hydrogen contains \(91.27 \% \mathrm{E}\) and \(8.73 \% \mathrm{H}\) by mass. If the formula of the compound is \(\mathrm{E}_{3} \mathrm{H}_{8}\), calculate the atomic mass of \(\mathrm{E}\).

Short answer

The atomic mass of the unknown element E is approximately 28.099 g/mol.

Step-by-step solution

Calculate the mass of hydrogen in the compound

Since we know that hydrogen makes up 8.73% of the compound's mass, and we also know the formula is E3H8, we can calculate the total mass of hydrogen in the compound. From the periodic table, the atomic mass of hydrogen is 1.008 g/mol. We have 8 hydrogen atoms in one molecule of the compound, so the total mass of hydrogen in the compound is: 8 * 1.008 g/mol = 8.064 g/mol

Calculate the mass of E in the compound

We know that element E makes up 91.27% of the compound's mass. Also, we know that the total mass of hydrogen in the compound is 8.064 g/mol. Therefore, we can find the mass of the unknown element E in the compound by subtracting the mass of hydrogen from the total mass of the compound: Total mass of the compound = mass of E + mass of hydrogen let x be the mass of the compound and y be the mass of E, then we get \(y = x - 8.064\) Since we know the compound is 91.27% E and 8.73% hydrogen, this means that \(\frac{y}{x} = 0.9127\), or \(y = 0.9127x\).

Solve for E's mass in the compound

Now we have two equations with two variables (x and y). We can solve these equations to find the mass of E in the compound: \(y = 0.9127x\) \(y = x - 8.064\) Substituting the first equation into the second equation, we get: \(0.9127x = x - 8.064\) Rearrange the equation to isolate x: \(x - 0.9127x = 8.064\) \(0.0873x = 8.064\) Now, solve for x: \(x = \frac{8.064}{0.0873} \approx 92.36 \ g/mol\).

Calculate atomic mass of E

Now that we have found the mass of the compound, we can find the mass of E by substituting x back into one of the equations (using the first one): \(y = 0.9127x\) \(y = 0.9127 (92.36)\) \(y \approx 84.296\ g/mol\). However, there are 3 atoms of element E in the formula (E3H8), so we need to divide the mass of E in the compound by 3 to find the atomic mass of the element E: Atomic mass of E \(= \frac{84.296}{3}\) Atomic mass of E \(\approx 28.099\ g/mol\). So the atomic mass of the element E is approximately 28.099 g/mol.

Key concepts

Binary Compounds

A binary compound is a chemical compound composed of exactly two different elements. These two elements can either be metals and non-metals or two non-metals. The importance of understanding binary compounds lies in their simplicity, which makes it easier to study chemical relationships and reactions.
In the exercise, the binary compound is composed of an unknown element, E, and hydrogen (H). This means we have a combination of exactly two elements forming the compound. Binary compounds often have simple formulas that reflect the ratio of the atoms involved, such as E extsubscript{3}H extsubscript{8} in this case.
Understanding binary compounds allows students to explore basic chemical principles such as naming conventions and formula calculations. For binary compounds, the naming typically involves the first element keeping its name and the second element adopting an 'ide' suffix. This understanding helps in appreciating how elements interact chemically to form new substances.

Atomic Mass

Atomic mass, also known as atomic weight, is the average mass of atoms of an element, measured in atomic mass units (amu). It considers the mass and relative abundance of each isotope of the element. The atomic mass gives insight into the proportions of elements in chemical reactions, ensuring correct stoichiometric balances.
In our exercise, we calculated the atomic mass of the element E. Initially, we find out the total mass of the compound and use the percentage composition to isolate the mass of E within the compound. It then involves dividing this mass by the number of atoms of E present (which is 3 in the formula E extsubscript{3}H extsubscript{8}) to find its atomic mass.
The calculation ensures that students understand how to use data from the periodic table and formulae to derive theoretical atomic masses. Remember, having a grasp of atomic mass helps in predicting the outcomes of chemical reactions and understanding the nature of mixtures.

Percentage Composition

Percentage composition refers to the percentage by mass of each elemental component in a compound. It provides a quick insight into the proportions of each element present in a compound. This concept is vital for understanding the formula mass or molar mass of entire compounds.

• For example, in this exercise, the compound comprises 91.27% element E and 8.73% hydrogen by mass.
• This information allows us to derive the total mass of each element in a given amount of the compound and to find out the proportions used to establish the chemical formula.
• The process involves identifying how much of each element exists in the compound if we consider a specific sample mass or molar mass.

Knowing how to calculate percentage composition enables students to analyze and compare different substances accurately, leading to a broader understanding of chemical substance makeup.

Chemical Formula

A chemical formula represents the chemical composition of a compound using element symbols and numerical subscripts. These formulas provide a standard way of describing the types and numbers of atoms in the simplest form of a substance.
In this exercise, the given formula is E extsubscript{3}H extsubscript{8}, meaning there are three atoms of element E and eight hydrogen atoms. The formula not only informs us about the stoichiometry of elements in the compound, but also guides the calculation of its molecular mass.
The chemical formula is essential in studying reactions, particularly in understanding how reactants convert into products. It also supports learning about the conservation of mass and charge during chemical reactions.
Recognizing and utilizing chemical formulas allow students to predict the characteristics and reactivity of compounds, making it a fundamental aspect of chemistry education.