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What is the empirical formula of the ingredient in Bufferin tablets that has the percent composition \(14.25 \%\) C, \(56.93 \% \mathrm{O}\), and \(28.83 \% \mathrm{Mg}\) by mass?

Short Answer

Expert verified
The empirical formula is \( \text{MgCO}_3 \).

Step by step solution

01

Determine the Molar Masses

First, we need to find the molar masses of each element: Carbon (C) is approximately 12.01 g/mol, Oxygen (O) is approximately 16.00 g/mol, and Magnesium (Mg) is approximately 24.31 g/mol.
02

Convert Percentage to Grams

Assume we have a 100 g sample of the compound. For each element, convert the percentage to grams: \( 14.25 \) g of C, \( 56.93 \) g of O, and \( 28.83 \) g of Mg.
03

Convert Grams to Moles

Convert the grams of each element to moles by dividing by their molar masses: For C: \( \frac{14.25}{12.01} \approx 1.19 \) moles. For O: \( \frac{56.93}{16.00} \approx 3.56 \) moles. For Mg: \( \frac{28.83}{24.31} \approx 1.19 \) moles.
04

Determine the Simplest Whole Number Ratio

Divide the moles of each element by the smallest number of moles calculated: For all elements, divide by 1.19 moles. This gives C: 1, O: 3, Mg: 1.
05

Formulate the Empirical Formula

Write the empirical formula using the whole number ratios determined: The empirical formula is \( \text{MgCO}_3 \).

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Key Concepts

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

Percent Composition
Percent composition is a way to express how much of each element is present in a chemical compound by mass. Imagine breaking down a compound into its elements and figuring out what fraction of the total mass each element contributes. For instance, if a compound contains 14.25% carbon, this means that carbon makes up 14.25 grams out of every 100 grams of the compound. This is essential because it helps chemists understand the makeup of the substance, which is the first step in identifying an empirical formula.
To convert percent composition to a usable form for further calculations, it is common to imagine having 100 grams of the compound. This makes it very straightforward: the percent value becomes the mass in grams for that element. This assumption simplifies moving on to the next steps, such as determining the ratio of elements, which leads to discovering the compound's empirical formula. So percent composition acts as a bridge from a compound's mass data to its formula.
Molar Mass
Molar mass is the mass of one mole of a given element or compound, expressed in grams per mole (g/mol). Each element in the periodic table has a specific molar mass, which is typically a number around the atomic weight of the element. For example, a mole of carbon weighs about 12.01 g/mol, while a mole of oxygen weighs about 16.00 g/mol.
The concept of molar mass is crucial in converting between grams and moles, which is often necessary to solve chemical problems. Knowing the molar mass allows us to convert from the mass of each element in a compound to the number of moles. This step bridges the gap between the laboratory (where masses are measured) and chemistry calculations (where moles are needed).
When determining an empirical formula, the molar masses of the elements in the compound help compute the amount of each element in moles. It's the foundation for comparing quantities of different elements, which helps in determining the simplest ratio between them.
Moles
The mole is a fundamental unit in chemistry that represents a quantity of chemical entities, usually atoms or molecules, and is equivalent to Avogadro's number, roughly 6.022 x 10²³ particles. Moles provide a way to count these entities in a manageable and meaningful way since they are too small to count individually.
The conversion between the mass of a substance and the amount in moles is central in solving chemical problems. By dividing the mass of an element by its molar mass, you find out how many moles of that element are present. For example, when determining an empirical formula, this calculation reveals how many moles of each element are in a sample.
After converting to moles, it's possible to establish the simplest ratio of each element in a compound. This whole number ratio is what gives us the empirical formula. The mole is a unit that facilitates this crucial step in the analysis.
Chemical Compound
A chemical compound is a substance composed of two or more different elements that are chemically bonded together. The way elements combine and the ratio of these combinations define the unique characteristics of each compound. A compound's properties greatly depend on these combinations and arrangements.
For a compound like the ingredient in Bufferin tablets, we can use information such as the percent composition and the results from calculations involving molar mass and moles to find the compound's empirical formula. This empirical formula, such as \( \text{MgCO}_3 \), represents the simplest whole-number ratio of elements in the compound. It's a crucial form of notation that chemists use to describe compounds without specifying the actual number of atoms or their arrangement, which would be detailed in a molecular formula.
The empirical formula provides the foundation for understanding the compound's nature and its possible reactions since it identifies the elements and their proportions, which is integral to applied chemistry.

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Most popular questions from this chapter

Sodium azide \(\left(\mathrm{NaN}_{3}\right)\) yields \(\mathrm{N}_{2}\) gas when heated to \(300{ }^{\circ} \mathrm{C}\), a reaction used in automobile air bags. If \(1.00 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) has a volume of \(47.0 \mathrm{~L}\) under the reaction conditions, how many liters of gas can be formed by heating \(38.5 \mathrm{~g}\) of \(\mathrm{NaN}_{3} ?\) The reaction is $$ 2 \mathrm{NaN}_{3}(s) \longrightarrow 3 \mathrm{~N}_{2}(g)+2 \mathrm{Na}(s) $$

When eaten, dietary carbohydrates are digested to yield glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\), which is then metabolized to yield carbon dioxide and water: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} $$ Unbalanced Balance the equation, and calculate both the mass in grams and the volume in liters of the \(\mathrm{CO}_{2}\) produced from \(66.3 \mathrm{~g}\) of glucose, assuming that \(1 \mathrm{~mol}\) of \(\mathrm{CO}_{2}\) has a volume of \(25.4 \mathrm{~L}\) at normal body temperature.

A compound of formula \(\mathrm{XCl}_{3}\) reacts with aqueous \(\mathrm{AgNO}_{3}\) to yield solid AgCl according to the following equation: $$ \mathrm{XCl}_{3}(a q)+3 \mathrm{AgNO}_{3}(a q) \longrightarrow \mathrm{X}\left(\mathrm{NO}_{3}\right)_{3}(a q)+3 \mathrm{AgCl}(s) $$ When a solution containing \(0.634 \mathrm{~g}\) of \(\mathrm{XCl}_{3}\) was allowed to react with an excess of aqueous \(\mathrm{AgNO}_{3}, 1.68 \mathrm{~g}\) of solid \(\mathrm{AgCl}\) was formed. What is the identity of the atom \(\mathrm{X} ?\)

Compound X contains only carbon, hydrogen, nitrogen, and chlorine. When \(1.00 \mathrm{~g}\) of \(\mathrm{X}\) is dissolved in water and allowed to react with excess silver nitrate, \(\mathrm{AgNO}_{3}\), all the chlorine in \(\mathrm{X}\) reacts and \(1.95 \mathrm{~g}\) of solid \(\mathrm{AgCl}\) is formed. When \(1.00 \mathrm{~g}\) of \(\mathrm{X}\) undergoes complete combustion, \(0.900 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.735 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are formed. What is the empirical formula of \(\mathrm{X}\) ?

The reaction of tungsten hexachloride \(\left(\mathrm{WCl}_{6}\right)\) with bismuth gives hexatungsten dodecachloride \(\left(\mathrm{W}_{6} \mathrm{Cl}_{12}\right)\) $$ \mathrm{WCl}_{6}+\mathrm{Bi} \longrightarrow \mathrm{W}_{6} \mathrm{Cl}_{12}+\mathrm{BiCl}_{3} \quad \text { Unbalanced } $$ (a) Balance the equation. (b) How many grams of bismuth react with \(150.0 \mathrm{~g}\) of \(\mathrm{WCl}_{6} ?\) (c) When \(228 \mathrm{~g}\) of \(\mathrm{WCl}_{6}\) react with \(175 \mathrm{~g}\) of \(\mathrm{Bi}\), how much \(\mathrm{W}_{6} \mathrm{Cl}_{12}\) is formed based on the limiting reactant?

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