Empirical Formula Of Iron Oxide

Learning Objectives

Past the end of this section, you volition be able to:

• Compute the percent composition of a compound
• Determine the empirical formula of a compound
• Determine the molecular formula of a compound

The previous section discussed the relationship betwixt the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, one may determine the corporeality of the substance (moles) from its mass, and vice versa. But what if the chemic formula of a substance is unknown? In this section, these aforementioned principles will exist applied to derive the chemical formulas of unknown substances from experimental mass measurements.

Percent Composition

The elemental makeup of a chemical compound defines its chemic identity, and chemic formulas are the most succinct manner of representing this elemental makeup. When a compound's formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound'south percent limerick, divers as the percentage by mass of each chemical element in the compound. For example, consider a gaseous compound equanimous solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

$$\%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%$$

$$\%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%$$

If analysis of a ten.0-g sample of this gas showed information technology to comprise 2.5 yard H and seven.5 g C, the percent composition would be calculated to be 25% H and 75% C:

$$\%\text{H}=\frac{\mathrm{two.5}\text{g H}}{10.0\text{g compound}}\times 100\%=25\%$$

$$\%\text{C}=\frac{7.5\text{thousand C}}{10.0\text{grand compound}}\times 100\%=75\%$$

Case three.9

Calculation of Percent Composition

Analysis of a 12.04-g sample of a liquid compound equanimous of carbon, hydrogen, and nitrogen showed it to contain 7.34 one thousand C, 1.85 chiliad H, and ii.85 grand N. What is the percent limerick of this chemical compound?

Solution

To calculate percent limerick, divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:

$$\begin{array}{c}\\ \%\text{C}=\frac{7.34\text{k C}}{12.04\text{g compound}}\times 100\%=61.0\%\\ \%\text{H}=\frac{1.85\text{g H}}{12.04\text{thou compound}}\times 100\%=\mathrm{fifteen.iv}\%\\ \%\text{North}=\frac{2.85\text{g N}}{12.04\text{g compound}}\times 100\%=23.7\%\end{array}$$

The analysis results indicate that the compound is 61.0% C, 15.iv% H, and 23.7% Northward past mass.

Check Your Learning

A 24.81-yard sample of a gaseous compound containing but carbon, oxygen, and chlorine is determined to contain iii.01 g C, 4.00 one thousand O, and 17.81 g Cl. What is this compound's percent composition?

Answer:

12.i% C, 16.1% O, 71.79% Cl

Determining Percent Limerick from Molecular or Empirical Formulas

Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH₃), ammonium nitrate (NH₄NO₃), and urea (CH_ivN₂O). The chemical element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the chemical compound is a practical and economic business concern for consumers choosing among these fertilizers. For these sorts of applications, the percent limerick of a compound is easily derived from its formula mass and the atomic masses of its elective elements. A molecule of NH_iii contains i N cantlet weighing xiv.01 amu and three H atoms weighing a total of (three $\times$ 1.008 amu) = three.024 amu. The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percentage composition is:

$$\begin{array}{c}\\ \%\text{Northward}=\frac{14.01\text{amu N}}{17.03\text{amu}{\text{NH}}_3}\times 100\%=82.27\%\\ \%\text{H}=\frac{3.024\text{amu H}}{17.03\text{amu}{\text{NH}}_3}\times 100\%=17.76\%\end{array}$$

This same approach may exist taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is well-nigh convenient and would simply involve the use of molar masses instead of diminutive and formula masses, as demonstrated Example 3.10. Equally long as the molecular or empirical formula of the chemical compound in question is known, the percent composition may be derived from the atomic or molar masses of the compound's elements.

Example 3.10

Determining Percentage Composition from a Molecular Formula

Aspirin is a chemical compound with the molecular formula C_nineH₈O_four. What is its percent composition?

Solution

To calculate the percent limerick, the masses of C, H, and O in a known mass of C₉H₈O_four are needed. It is convenient to consider 1 mol of C_nineH₈O_four and use its molar mass (180.159 grand/mole, determined from the chemical formula) to calculate the percentages of each of its elements:

$$\begin{array}{c}\\ \%\text{C}=\frac{9\text{mol C}\times \text{tooth mass C}}{\text{tooth mass}{\text{C}}_9{\text{H}}_{\mathrm{eight}}{\text{O}}_{\mathrm{four}}}\times 100=\frac{9\times 12.01\text{m/mol}}{180.159\text{g/mol}}\times 100=\frac{108.09\text{g/mol}}{180.159\text{g/mol}}\times 100\\ \%\text{C}=60.00\%\text{C}\hfill \end{array}$$

$$\begin{array}{c}\\ \%\text{H}=\frac{\mathrm{eight}\text{mol H}\times \text{molar mass H}}{\text{molar mass}{\text{C}}_9{\text{H}}_8{\text{O}}_4}\times 100=\frac{8\times 1.008\text{chiliad/mol}}{180.159\text{m/mol}}\times 100=\frac{8.064\text{thousand/mol}}{180.159\text{g/mol}}\times 100\\ \%\text{H}=4.476\%\text{H}\hfill \end{array}$$

$$\begin{array}{c}\\ \%\text{O}=\frac{\mathrm{four}\text{mol O}\times \text{molar mass O}}{\text{molar mass}{\text{C}}_9{\text{H}}_{\mathrm{eight}}{\text{O}}_4}\times 100=\frac{4\times 16.00\text{g/mol}}{180.159\text{thou/mol}}\times 100=\frac{64.00\text{one thousand/mol}}{180.159\text{g/mol}}\times 100\\ \%\text{O}=35.52\%\hfill \end{array}$$

Notation that these percentages sum to equal 100.00% when appropriately rounded.

Cheque Your Learning

To three pregnant digits, what is the mass percentage of iron in the compound Fe_iiO₃?

Determination of Empirical Formulas

Every bit previously mentioned, the most common approach to determining a compound'due south chemic formula is to first measure the masses of its elective elements. Withal, go along in mind that chemical formulas represent the relative numbers, not masses, of atoms in the substance. Therefore, any experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound. This is accomplished using molar masses to catechumen the mass of each element to a number of moles. These tooth amounts are used to compute whole-number ratios that can be used to derive the empirical formula of the substance. Consider a sample of compound adamant to incorporate 1.71 g C and 0.287 chiliad H. The corresponding numbers of atoms (in moles) are:

$$\begin{array}{l}\\ 1.71\text{g C}\times \frac{1\text{mol C}}{12.01\text{m C}}=0.142\text{mol C}\\ 0.287\text{thousand H}\times \frac{1\text{mol H}}{1.008\text{grand H}}=0.284\text{mol H}\end{array}$$

Thus, this compound may be represented by the formula C_0.142H_0.284. Per convention, formulas contain whole-number subscripts, which tin can exist achieved by dividing each subscript past the smaller subscript:

$${\text{C}}_{\frac{0.142}{0.142}}{\text{H}}_{\frac{0.284}{0.142}}\text{or}{\text{CH}}_2$$

(Recall that subscripts of "one" are not written but rather causeless if no other number is nowadays.)

The empirical formula for this compound is thus CH₂. This may or may not exist the compound'south molecular formula also; however, additional information is needed to make that conclusion (as discussed later on in this section).

Consider every bit another example a sample of compound determined to contain v.31 g Cl and 8.twoscore g O. Post-obit the aforementioned approach yields a tentative empirical formula of:

$${\text{Cl}}_{0.150}{\text{O}}_{0.525}={\text{Cl}}_{\frac{0.150}{0.150}}{\text{O}}_{\frac{0.525}{0.150}}={\text{ClO}}_{3.5}$$

In this instance, dividing by the smallest subscript nevertheless leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, multiply each of the subscripts past ii, retaining the same cantlet ratio and yielding Cl₂O₇ as the final empirical formula.

In summary, empirical formulas are derived from experimentally measured chemical element masses by:

1. Deriving the number of moles of each chemical element from its mass
2. Dividing each element's molar corporeality by the smallest tooth amount to yield subscripts for a tentative empirical formula
3. Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained

Figure 3.11 outlines this procedure in menses chart manner for a substance containing elements A and X.

Figure 3.eleven The empirical formula of a chemical compound can be derived from the masses of all elements in the sample.

Instance three.11

Determining a Compound's Empirical Formula from the Masses of Its Elements

A sample of the blackness mineral hematite (Figure iii.12), an oxide of iron found in many fe ores, contains 34.97 thou of atomic number 26 and fifteen.03 g of oxygen. What is the empirical formula of hematite?

Effigy 3.12 Hematite is an iron oxide that is used in jewelry. (credit: Mauro Cateb)

Solution

This problem provides the mass in grams of each chemical element. Begin by finding the moles of each:

$$\begin{array}{l}\\ 34.97\text{g Atomic number 26}\left(\frac{\text{mol Fe}}{55.85\text{thousand}}\right)& =& 0.6261\text{mol Fe}\hfill \\ 15.03\text{1000 O}\left(\frac{\text{mol O}}{\mathrm{sixteen.00}\text{k}}\right)& =& 0.9394\text{mol O}\hfill \end{array}$$

Next, derive the iron-to-oxygen molar ratio past dividing past the bottom number of moles:

$$\begin{array}{}\\ \\ \frac{0.6261}{0.6261}=1.000\text{mol Fe}\\ \frac{0.9394}{0.6261}=\mathrm{i.500}\text{mol O}\end{array}$$

The ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe₁O_i.5). Finally, multiply the ratio past two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio:

$$2\left({\text{Fe}}_{\mathrm{ane}}{\text{O}}_{1.5}\right)={\text{Fe}}_2{\text{O}}_{\mathrm{iii}}$$

The empirical formula is Atomic number 26₂O₃.

Check Your Learning

What is the empirical formula of a chemical compound if a sample contains 0.130 g of nitrogen and 0.370 g of oxygen?

Link to Learning

For boosted worked examples illustrating the derivation of empirical formulas, lookout man the brief video clip.

Deriving Empirical Formulas from Percent Composition

Finally, with regard to deriving empirical formulas, consider instances in which a chemical compound'due south percent composition is available rather than the absolute masses of the compound's constituent elements. In such cases, the percent composition can be used to calculate the masses of elements nowadays in any convenient mass of compound; these masses tin and so exist used to derive the empirical formula in the usual manner.

Case iii.12

Determining an Empirical Formula from Pct Limerick

The bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29% C and 72.71% O (Figure 3.xiii). What is the empirical formula for this gas?

Figure three.13 An oxide of carbon is removed from these fermentation tanks through the large copper pipes at the top. (credit: "Dual Freq"/Wikimedia Commons)

Solution

Since the scale for percentages is 100, information technology is most convenient to summate the mass of elements present in a sample weighing 100 thousand. The adding is "nigh convenient" because, per the definition for pct limerick, the mass of a given element in grams is numerically equivalent to the chemical element'due south mass percentage. This numerical equivalence results from the definition of the "percentage" unit, whose name is derived from the Latin phrase per centum meaning "by the hundred." Considering this definition, the mass percentages provided may be more conveniently expressed as fractions:

$$\begin{array}{l}\\ 27.29\%\text{C}& =\hfill & \frac{27.29\text{g C}}{100\text{chiliad compound}}\hfill \\ 72.71\%\text{O}& =\hfill & \frac{72.71\text{g O}}{100\text{1000 compound}}\hfill \end{array}$$

The molar amounts of carbon and oxygen in a 100-g sample are calculated past dividing each chemical element's mass by its molar mass:

$$\begin{array}{l}\\ 27.29\text{thousand C}\left(\frac{\text{mol C}}{12.01\text{1000}}\right)& =\hfill & 2.272\text{mol C}\hfill \\ 72.71\text{yard O}\left(\frac{\text{mol O}}{16.00\text{k}}\right)& =\hfill & \mathrm{iv.544}\text{mol O}\hfill \end{array}$$

Coefficients for the tentative empirical formula are derived by dividing each molar corporeality by the lesser of the two:

$$\begin{array}{}\\ \\ \frac{2.272\text{mol C}}{2.272}=1\\ \frac{4.544\text{mol O}}{\mathrm{ii.272}}=2\end{array}$$

Since the resulting ratio is one carbon to ii oxygen atoms, the empirical formula is CO₂.

Check Your Learning

What is the empirical formula of a compound containing twoscore.0% C, vi.71% H, and 53.28% O?

Derivation of Molecular Formulas

Recall that empirical formulas are symbols representing the relative numbers of a compound's elements. Determining the absolute numbers of atoms that compose a unmarried molecule of a covalent compound requires knowledge of both its empirical formula and its molecular mass or molar mass. These quantities may be determined experimentally past various measurement techniques. Molecular mass, for case, is ofttimes derived from the mass spectrum of the chemical compound (see discussion of this technique in the previous chapter on atoms and molecules). Molar mass tin exist measured by a number of experimental methods, many of which will be introduced in afterward chapters of this text.

Molecular formulas are derived by comparing the chemical compound'south molecular or molar mass to its empirical formula mass. As the name suggests, an empirical formula mass is the sum of the average diminutive masses of all the atoms represented in an empirical formula. If the molecular (or tooth) mass of the substance is known, information technology may be divided by the empirical formula mass to yield the number of empirical formula units per molecule (n):

$$\frac{\text{molecular or molar mass}\left(\text{amu or}\frac{\text{one thousand}}{\text{mol}}\right)}{\text{empirical formula mass}\left(\text{amu or}\frac{\text{g}}{\text{mol}}\right)}=\mathrm{northward}\text{formula units/molecule}$$

The molecular formula is then obtained past multiplying each subscript in the empirical formula by n, every bit shown by the generic empirical formula A_xB_y:

$${({\text{A}}_{\text{x}}{\text{B}}_{\text{y}})}_{\text{n}}={\text{A}}_{\text{nx}}{\text{B}}_{\text{ny}}$$

For instance, consider a covalent chemical compound whose empirical formula is determined to be CH₂O. The empirical formula mass for this chemical compound is approximately 30 amu (the sum of 12 amu for one C atom, ii amu for two H atoms, and 16 amu for one O atom). If the compound'south molecular mass is adamant to be 180 amu, this indicates that molecules of this compound contain half dozen times the number of atoms represented in the empirical formula:

$$\frac{180\text{amu/molecule}}{\mathrm{thirty}\frac{\text{amu}}{\text{formula unit}}}=6\text{formula units/molecule}$$

Molecules of this compound are and then represented past molecular formulas whose subscripts are six times greater than those in the empirical formula:

$${\text{(CH}}_{\text{2}}{\text{O)}}_{\text{half-dozen}}={\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$$

Note that this same approach may be used when the tooth mass (g/mol) instead of the molecular mass (amu) is used. In this case, one mole of empirical formula units and molecules is considered, every bit opposed to single units and molecules.

Example 3.xiii

Determination of the Molecular Formula for Nicotine

Nicotine, an alkaloid in the nightshade family unit of plants that is mainly responsible for the addictive nature of cigarettes, contains 74.02% C, viii.710% H, and 17.27% N. If 40.57 yard of nicotine contains 0.2500 mol nicotine, what is the molecular formula?

Solution

Determining the molecular formula from the provided data will crave comparison of the compound's empirical formula mass to its molar mass. Every bit the first footstep, use the percent composition to derive the chemical compound'southward empirical formula. Assuming a convenient, a 100-g sample of nicotine yields the post-obit molar amounts of its elements:

$$\begin{array}{l}\\ \left(74.02\text{g C}\right)\left(\frac{\mathrm{i}\text{mol C}}{12.01\text{g C}}\right)& =\hfill & \mathrm{half\; dozen.163}\text{mol C}\hfill \\ \left(8.710\text{m H}\right)\left(\frac{\mathrm{one}\text{mol H}}{\mathrm{ane.01}\text{g H}}\right)& =\hfill & \mathrm{viii.624}\text{mol H}\hfill \\ \left(17.27\text{g N}\right)\left(\frac{\mathrm{i}\text{mol N}}{\mathrm{fourteen.01}\text{m N}}\right)& =\hfill & 1.233\text{mol N}\hfill \end{array}$$

Next, calculate the tooth ratios of these elements relative to the least arable element, North.

$$\mathrm{half\; dozen.163}\text{mol}C/1.233\text{mol}\mathrm{Due\; north}=\mathrm{five}$$

$$8.264\text{mol}H/\mathrm{ane.233}\text{mol}\mathrm{N}=\mathrm{vii}$$

$$1.233\text{mol}N/1.233\text{mol}\mathrm{N}=1$$

$$\begin{array}{lll}\frac{\mathrm{one.233}}{1.233}& =\hfill & 1.000\text{mol Northward}\hfill \\ \frac{6.163}{1.233}& =\hfill & 4.998\text{mol C}\hfill \\ \frac{\mathrm{viii.624}}{\mathrm{ane.233}}& =\hfill & \mathrm{six.994}\text{mol H}\hfill \end{array}$$

The C-to-N and H-to-N tooth ratios are adequately shut to whole numbers, and so the empirical formula is C₅H_viiN. The empirical formula mass for this chemical compound is therefore 81.xiii amu/formula unit, or 81.13 m/mol formula unit.

Calculate the tooth mass for nicotine from the given mass and molar corporeality of compound:

$$\frac{40.57\text{k nicotine}}{0.2500\text{mol nicotine}}=\frac{162.3\text{g}}{\text{mol}}$$

Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:

$$\frac{\mathrm{162.three}\text{m/mol}}{81.13\frac{\text{grand}}{\text{formula unit}}}=2\text{formula units/molecule}$$

Finally, derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by ii:

$${\text{(C}}_{\text{five}}{\text{H}}_{\text{7}}{\text{North)}}_{\text{two}}={\text{C}}_{\text{ten}}{\text{H}}_{\text{14}}{\text{Northward}}_{\text{2}}$$

Check Your Learning

What is the molecular formula of a chemical compound with a pct limerick of 49.47% C, 5.201% H, 28.84% N, and xvi.48% O, and a molecular mass of 194.two amu?

Empirical Formula Of Iron Oxide,

Source: https://openstax.org/books/chemistry-2e/pages/3-2-determining-empirical-and-molecular-formulas

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