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Chapter 3: Problem 49

A compound whose empirical formula is \(\mathrm{XF}_{3}\) consists of \(65 \%\) \(\mathrm{F}\) by mass. What is the atomic mass of \(\mathrm{X} ?\)

Short Answer

Expert verified
The atomic mass of element X in the compound XF3, which is 65% fluorine by mass, is approximately \(30.69 g/mol\).

Step by step solution

01

Determine the atomic mass of fluorine and calculate the mass of 3 fluorine atoms in the compound

The atomic mass of fluorine (F) is approximately 19 g/mol. Since there are 3 fluorine atoms in the formula XF3, we can calculate the mass of fluorine in the compound: Mass of 3 F atoms = 3 脳 atomic mass of F = 3 脳 19 g/mol = 57 g/mol
02

Calculate the mass of element X in the compound

Since the compound is 65% fluorine by mass, the remaining percentage must be the mass of element X in the compound. Therefore, the mass percentage of element X in the compound is: 100% - 65% = 35% Now, we can determine the mass of element X in the compound by finding the mass equivalent of 35%. Since the mass of 3 F atoms in the compound equals 57 g/mol (calculated in Step 1), we can set up the equation: (35 g X) / (57 g F) = 35% Solve for the mass of X: 35 g X = 0.35 脳 57 g F Mass of X = 19.95 g/mol Since the mass percentage of a single atom of X should be the same as the mass percentage in the compound, we can set up the equation: Mass of 1 X atom / (57 g F + Mass of 1 X atom) = 35%
03

Calculate the atomic mass of element X

Now, we will solve the equation we set up in Step 2 to find the atomic mass of element X: Mass of 1 X atom / (57 g F + Mass of 1 X atom) = 35% Let the atomic mass of X be 'm'. m / (57 + m) = 0.35 Solving for 'm', we get: m = 0.35 脳 (57 + m) m = 19.95 + 0.35m Now, subtract 0.35m from both sides to get: 0.65m = 19.95 Finally, we can find the atomic mass of X by dividing both sides by 0.65: m = 19.95 / 0.65 m 鈮� 30.69 g/mol Thus, the atomic mass of element X is approximately 30.69 g/mol.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Empirical Formula
The empirical formula is a crucial concept in chemistry and represents the simplest ratio of elements in a compound. For instance, the empirical formula of the compound in the exercise is \(\mathrm{XF}_{3}\). This signifies that for every atom of element \(X\), there are three atoms of fluorine.

The empirical formula does not necessarily convey the exact number of each type of atom in the compound. It only shows the lowest possible ratio. However, it provides foundational information necessary to determine the compound's molecular formula, if the molar mass is known.
  • Empirical formulas simplify comparison between different compounds.
  • Helps in calculating relative molar masses.
  • Often used as a starting point for finding molecular formulas.
Fluorine
Fluorine is an essential element in chemistry, especially in organofluorine compounds. In the problem provided, fluorine's atomic mass is noted as 19 g/mol, and it plays a significant role in determining the mass percentage composition of the compound \(\mathrm{XF}_{3}\).

Fluorine is highly electronegative, affecting the properties of compounds it forms. Its presence in a compound can influence both its chemical reactivity and stability. Calculating the mass of 3 fluorine atoms involved a multiplication of its atomic mass by 3, which helped in setting the groundwork for further calculations in this problem.
  • Commonly used in halogenation reactions.
  • A major component in non-stick coatings and refrigerants.
  • Presents in high percentages, influences compound's properties significantly.
Mass Percentage Composition
Understanding mass percentage composition is vital for determining the relative amounts of components in a compound. In our exercise, around 65% of the compound is made up of fluorine. This means that when you calculate the mass of component elements, you attribute 65% of the total mass to fluorine.

Mass percentage is calculated by dividing the mass of the element of interest by the total mass of the compound, and then multiplying the result by 100. In this problem:
  • Knowing fluorine's contribution helped determine the remaining mass percentage for element X.
  • Essential in calculating unknown component鈥檚 proportional weight in a compound.
  • Useful in various applications, including formulation of medicines and materials.
By understanding and calculating mass percentage composition, we can clearly see the distribution of each element in the compound, which plays a crucial role in determining molecular and empirical formulas.

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Most popular questions from this chapter

The allowable concentration level of vinyl chloride, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl},\) in the atmosphere in a chemical plant is \(2.0 \times 10^{-6} \mathrm{~g} / \mathrm{L}\). How many moles of vinyl chloride in each liter does this represent? How many molecules per liter?

When benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) reacts with bromine \(\left(\mathrm{Br}_{2}\right),\) bromobenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\right)\) is obtained: $$ \mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{Br}_{2} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}+\mathrm{HBr} $$ (a) When \(30.0 \mathrm{~g}\) of benzene reacts with \(65.0 \mathrm{~g}\) of bromine, what is the theoretical yield of bromobenzene? (b) If the actual yield of bromobenzene is \(42.3 \mathrm{~g}\), what is the percentage yield?

Hydrofluoric acid, HF \((a q)\), cannot be stored in glass bottles because compounds called silicates in the glass are attacked by the \(\mathrm{HF}(a q)\). Sodium silicate \(\left(\mathrm{Na}_{2} \mathrm{SiO}_{3}\right),\) for example, reacts as follows: $$ \mathrm{Na}_{2} \mathrm{SiO}_{3}(s)+8 \mathrm{HF}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{SiF}_{6}(a q)+2 \mathrm{NaF}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) How many moles of HF are needed to react with 0.300 mol of \(\mathrm{Na}_{2} \mathrm{SiO}_{3} ?\) (b) How many grams of NaF form when \(0.500 \mathrm{~mol}\) of HF reacts with excess \(\mathrm{Na}_{2} \mathrm{SiO}_{3} ?\) (c) How many grams of \(\mathrm{Na}_{2} \mathrm{SiO}_{3}\) can react with \(0.800 \mathrm{~g}\) of HF?

Write balanced chemical equations to correspond to each of the following descriptions: (a) When sulfur trioxide gas reacts with water, a solution of sulfuric acid forms. (b) Boron sulfide, \(\mathrm{B}_{2} \mathrm{~S}_{3}(s),\) reacts violently with water to form dissolved boric acid, \(\mathrm{H}_{3} \mathrm{BO}_{3},\) and hydrogen sulfide gas. (c) Phosphine, \(\mathrm{PH}_{3}(g)\), combusts in oxygen gas to form water vapor and solid tetraphosphorus decaoxide. (d) When solid mercury(II) nitrate is heated, it decomposes to form solid mercury(II) oxide, gaseous nitrogen dioxide, and oxygen. (e) Copper metal reacts with hot concentrated sulfuric acid solution to form aqueous copper(II) sulfate, sulfur dioxide gas, and water.

When hydrocarbons are burned in a limited amount of air, both \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) form. When \(0.450 \mathrm{~g}\) of a particular hydrocarbon was burned in air, \(0.467 \mathrm{~g}\) of \(\mathrm{CO}, 0.733 \mathrm{~g}\) of \(\mathrm{CO}_{2},\) and \(0.450 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) were formed. (a) What is the empirical formula of the compound? (b) How many grams of \(\mathrm{O}_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?
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