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Considering your answer to Exercise \(81,\) which type of formula, empirical or molecular, can be obtained from elemental analysis that gives percent composition?

Short Answer

Expert verified
Only an empirical formula can be obtained from elemental analysis that gives percent composition, as it provides the simplest whole number ratio of elements present in a compound. The molecular formula cannot be directly determined from percent composition data as it requires information on the actual number of atoms of each element in the compound.

Step by step solution

01

Understand empirical and molecular formulas

Empirical formulas are the simplest whole number ratio of elements present in a compound, while molecular formulas are the actual number of atoms of each element present in a compound. The molecular formula is always a multiple of the empirical formula. For example, the empirical formula of hydrogen peroxide is HO while its molecular formula is H2O2 (which is made up of two empirical formula units).
02

Understand elemental analysis and percent composition data

Elemental analysis is a method used to determine the relative amounts of different elements in a compound. The results are usually given in the form of a percent composition, which indicates the percentage of each element present. Percent composition can be determined using the mass of each element present in the compound and the total mass of the compound. The formula to find the percent composition of an element in a compound is: Percent composition =\(\frac{mass\,of\,element\phantom{}in\phantom{}the\phantom{}compound }{total\,mass\,of\,the\,compound}\) x 100
03

Determine which formula can be obtained from percent composition data

With percent composition data, one can initially determine the simplest whole number ratio of elements present in a compound, which is the empirical formula. However, since the molecular formula is a multiple of the empirical formula and we don't have information on the actual number of atoms of each element in the compound, we cannot directly determine the molecular formula from percent composition data. Therefore, only an empirical formula can be obtained from elemental analysis that gives percent composition.

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Key Concepts

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

Molecular Formula
A molecular formula represents the actual number of atoms of each element in a compound. It is like a detailed map that shows the complete details of the structure of the compound. This means it provides the exact count of every type of atom present in a molecule. The molecular formula might look complex at first. But, it is just an extension of the simplest form of the compound you may encounter, the empirical formula.

To understand it better, think about water. While the empirical formula gives the simplest atom ratio, the molecular formula for water is \( H_2O \) indicating there are two hydrogen atoms bonded to a single oxygen atom.

Sometimes, the molecular formula and empirical formula are the same, like in the case of water. However, for compounds like glucose \( C_6H_{12}O_6 \), the molecular formula includes the exact number of atoms, whereas the empirical formula would be \( CH_2O \), representing the simplest ratio by dividing each atom count by 6.

It's key to remember: without additional data, like molar mass, direct determination of the molecular formula from percentage composition isn't possible. Yet, it helps kickstart the process in identifying possibilities.
Elemental Analysis
Elemental analysis is an essential analytical technique used in chemistry to determine the elemental composition of a compound. Essentially, it tells us what mix of elements is there in the compound and their relative amounts.

This is done by measuring the mass of every element in the compound and comparing this to the total mass. The analysis provides crucial insights, helping chemists to understand and confirm the composition before moving to more involved analyses, like structural determination.

In practical terms, elemental analysis results are often given as percent composition. This breaks down into the percentage of each different element within the compound. For example, a common elemental analysis might reveal that a sample consists of 12% carbon, 88% oxygen, indicating high levels of oxygen relative to carbon.

While elemental analysis is thorough, the process itself only unveils an empirical formula, not a molecular one. This limitation stems from lacking the mass needed to precisely determine how many times the empirical formula unit repeats within the molecule.
Percent Composition
Percent composition is a handy way to express the amount of each element in a compound, stated as a percentage of the total mass. This concept reflects the mass ratio of each element to the overall mass of the compound, providing a straightforward understanding of its makeup.

Calculating percent composition involves a simple formula: first, determine the mass of each individual element in the compound, then divide by the total compound mass and multiply by 100. For example:
  • Mass of the element = 16 grams
  • Total mass of the compound = 50 grams
The percent composition would be \( \left(\frac{16}{50}\right) \times 100 \approx 32\% \).

This method serves as a foundation for deducing an empirical formula—the simplest ratio of elements in the compound. However, keep in mind that percent composition lacks the specific numerical detail required for the molecular formula, as it does not account for how these ratios play out in larger molecule structures. Still, it provides an essential stepping stone for theoretical deductions and further empirical analyses.

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Most popular questions from this chapter

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) plus other impurities. A 752 -g sample of impure iron ore is heated with excess carbon, producing 453 g of pure iron by the following reaction: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}(g) $$ What is the mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample? Assume that \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is the only source of iron and that the reaction is 100\(\%\) efficient.

Zinc and magnesium metal each reacts with hydrochloric acid to make chloride salts of the respective metals, and hydrogen gas. A 10.00 -g mixture of zinc and magnesium produces 0.5171 g of hydrogen gas upon being mixed with an excess of hydrochloric acid. Determine the percent magnesium by mass in the original mixture.

Acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\right)\) is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. $$ 2 \mathrm{C}_{3} \mathrm{H}_{6}(g)+2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ If \(15.0 \mathrm{g} \mathrm{C}_{3} \mathrm{H}_{6}, 10.0 \mathrm{g} \mathrm{O}_{2},\) and 5.00 \(\mathrm{g} \mathrm{NH}_{3}\) are reacted, what mass of acrylonitrile can be produced, assuming 100\(\%\) yield?

A common demonstration in chemistry courses involves adding a tiny speck of manganese(IV) oxide to a concentrated hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)\) solution. Hydrogen peroxide decomposes quite spectacularly under these conditions to produce oxygen gas and steam (water vapor). Manganese(IV) oxide is a catalyst for the decomposition of hydrogen peroxide and is not consumed in the reaction. Write the balanced equation for the decomposition reaction of hydrogen peroxide.

Glass is a mixture of several compounds, but a major constituent of most glass is calcium silicate, \(\mathrm{CaSiO}_{3}\) . Glass can be etched by treatment with hydrofluoric acid; HF attacks the calcium silicate of the glass, producing gaseous and water-soluble products (which can be removed by washing the glass). For example, the volumetric glassware in chemistry laboratories is often graduated by using this process. Balance the following equation for the reaction of hydrofluoric acid with calcium silicate. $$ \mathrm{CaSiO}_{3}(s)+\mathrm{HF}(a q) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$

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