Chemical formulas and composition stoichiometry
Stoichiometry
The word “stoichiometry” is derived from the Greek stoicheion, which means “element,” and metron, which means “measure.”
Stoichiometry describes the quantitative relationships among elements in compounds (composition stoichiometry) and among substances as they undergo chemical changes (reaction stoichiometry).
Main division of chemical compounds: organic vs inorganic Many of the molecules found in nature are organic compounds. • Organic compounds contain C-C or C-H bonds or both, often in combination with nitrogen, oxygen, sulfur, and other elements. • All of the other compounds are inorganic compounds. Name
Formula
Name
Formula
Name
Formula
water
H2O
sulfur dioxide
SO2
butane
C4H10
hydrogen peroxide
H2O2
sulfur trioxide
SO3
pentane
C5H12
hydrogen chloride*
HCl
carbon monoxide
CO
benzene
C6H6
sulfuric acid
H2SO4
carbon dioxide
CO2
methanol (methyl alcohol)
CH3OH
nitric acid
HNO3
methane
CH4
ethanol (ethyl alcohol)
CH3CH2OH
acetic acid
CH3COOH
ethane
C2H6
acetone
CH3COCH3
ammonia
NH3
propane
C3H8
diethyl ether (ether)
CH3CH2OCH2CH3
Chemical formulas The chemical formula gives the number of atoms of each type in the molecule. But this formula does not express the order in which the atoms in the molecules are bonded together. For instance, the structural formula of propane is C3H8 and shows that the molecule is composed of 3 carbons and 8 hydrogens.
The structural formula shows the order in which atoms are connected. The lines connecting atomic symbols represent chemical bonds between atoms The structural formula of propane shows that the three C atoms are linked in a chain, with three H atoms bonded to each of the end C atoms and two H atoms bonded to the center C.
H
H
H
H
C
C
C
H
H
H
H
Chemists sometimes write out the formula in a way to better convey the connectivity information, e.g., CH3CH2CH3, which is a longer representation of the propane chemical formula.
Chemical formulas Each molecule of acetic acid, CH3COOH, contains 2 carbon atoms, 4 hydrogen atoms, and 2 oxygens. Writing it as CH3COOH (instead of C2H4O2) includes useful bonding and structural information.
H H
C H
OH C O
Law of Definite Proportions A compound can be decomposed by chemical means into simpler substances, always in the same ratio by mass. E.g. Water originates from the reaction between two H and one O. This leads to the ratio 1:8 which is the ratio of the masses of H and O in the molecule of H2O. AWH = 1,0079 amu = 1 AWO = 15,9994 amu = 16 One oxygen is sixteen times the mass of a H. Hence, in the molecule of water
H2
O
2 amu
16 amu
1
:
H2O FW = 18
8
Water is 11.1% hydrogen and 88.9% oxygen by mass.
Law of Definite Proportions Carbon dioxide (CO2) is 27.3% carbon and 72.7% oxygen by mass Calcium oxide (CaO) is 71.5% calcium and 28.5% oxygen by mass Calcium carbonate (CaCO3) is 40.1% calcium, 12.0% carbon, 47.9% oxygen by mass
Observations such as these on many pure compounds (as CaCO3) led Joseph-Louis Proust in1799 to the statement of the Law of Definite Proportions (also known as the Law of Constant Composition):
Different samples of a compound always contain the same elements in the same proportion by mass; this corresponds to atoms of these elements combined in fixed numerical ratios.
Percent composition and chemical formulas 2-7 Percent Composition and Formulas of Compounds
If the formula of aiscompound is known, its chemical composition be expressed as as mass) If the chemical formula known, its chemical composition can can be expressed ht (mass) the mass percent of each elementininthe the compound (percent composition). For example, the mass C H A P percent T E R 2 • Cof H Eeach M I C A Lelement F O R M U L A S A N D Ccompound O M P O S I T I O N (percent S T O I C H I O M composition). ETRY
he %.E.g.
one carbon dioxide molecule, CO2, contains one C atom and two O atoms. Percentage is the part divided by the whole times 100% (or simply parts per hundred), so we can represent the percent composition of carbon dioxide as follows:
2-7 Percent Composition and Formulas of Compounds One CO2, contains one C atom and two O atoms.
If the formula of C adivided compound is known, chemical can be expressed Percentage is the part by the whole times 100%,composition so12.0 we amu can represent the as mass of AW of its C %C 5 percent of each 3 100% 5 in the 3 100% 5 3 100%For 5 example, 27.3% C s)percent the mass element (percent composition). composition dioxide follows: mass of CO 2carbon MWasofcompound CO 2 44.0 amu one carbon dioxide molecule, CO2, contains one C atom and two O atoms. Percentage is ) 2 ( 16.0 amu massdivided of O by the whole 2 times 3 AW100% of O (or simply parts the part per hundred), so we can rep%O 5 3 100% 5 3 100% 5 3 100% 5 72.7% O mass of CO MW of CO 44.0 amu 2 resent the percent composition of carbon2 dioxide as follows: One mole of CO (44.0 g) contains one mole of C atoms (12.0 g) and two moles of O atoms
2 mass of C AW of C 12.0 amu therefore masses5in the preceding calculation. These %C 5 (32.0 g). We could 3 100% 5 have used these 3 100% 3 100% 5 27.3% C mass of CO MW of CO 44.0 amu 2 numbers are 2the same as the ones usedionly the units are different. In Example 2-12 we
base our calculation on AW one mole ( 16.0 amu ) masswill of O 23 of Orather than one 2molecule. %O 5 3 100% 5 3 100% 5 3 100% 5 72.7% O mass of CO MW of CO 44.0 amu 2 2 (44.0 g) contains one2 mole of C atoms (12.0 g) and two moles of O One mole of CO
atoms (32.0 g).
One mole of CO2 (44.0 g) contains one mole of C atoms (12.0 g) and two moles of O atoms (32.0 g). We could therefore have used these masses in the preceding calculation. These numbers are the same as the ones usedionly the units are different. In Example 2-12 we EXAMPLE 2-12 Percent Composition will base our calculation on one mole rather than one molecule. Calculate the percent composition by mass of HNO3.
Percent composition and chemical formulas Sample exercise • Calculate the percent composition by mass of HNO3
We first calculate the mass of one mole (molar mass). Then we express the mass of each element as a percent of the total. The molar mass of HNO3 is: Number of Mol of Atoms x Mass of One Mol of Atoms = Mass Due to Element 1x H =1 mol x 1,0 g =1,0 g of H 1x N =1 mol x 14,0 g = 14,0 g of N 3x O = 3 mol x 16,0 g = 48,0 g of O Mass of 1 mol of HNO Now, its percent composition is:
•% of H = (1.0 g /63,0 g) 100% = 1,6 % •% of N = (14.0 g /63,0 g) 100% = 22,2 % •% of O = (48,0 g /63,0 g) 100% = 76,2 % Remember The amount of substance that contains the mass in grams numerically equal to its formula weight in amu contains one mole of the substance. This is the molar mass and is numerically equal to the formula weight and has the units grams/mole (g/mol).
Percent composition and chemical formulas Practice exercise • Calculate the percent composition by mass of each of the following compounds Dopamine: C8H11NO2 Vitamin E: C29H50O2 Vanillin: C8H8O3
Derivation of Formulas from Composition The molecular formula indicates the numbers of atoms in a molecule. The empirical formula is the smallest whole-number ratio of atoms present. E.g. The empirical and molecular formulas for water are both H2O; for hydrogen peroxide, the empirical formula is HO, and the molecular formula is H2O2 Molecular formula
Simplest formula
C2H4O2 C6H12O6 H3PO4 H2O H2O2
CH2O CH2O H3PO4 H2O HO
Derivation of Formulas from Composition One of the first steps in characterizing a new compound is the determination of its percent composition.
A qualitative analysis is performed to determine which elements are present in the compound. A quantitative analysis is performed to determine the amount of each element. Once the percent composition of a compound is known, the empirical formula can be determined
CHAPTER 2 • CHEMICAL FORMULAS AND COMPOSITION STOICHIOMETRY Derivation of Formulas from Composition
m obtain mol x Fw Sample exercise Step 2: Now(Empirical we know thatFormula) 100.0 g of the compound contains 1.56 mol of S atoms and several 3.12 mol of O atoms. We obtain a whole-number ratio between these numbers ch • Compounds containing S and O are serious air pollutants and they represent the major
ber, and cause all the mass. llest ate
the ratioofofthis atoms in the simplest formula. of acidthat rain.gives A sample compound contains 50.1% sulfur and 49.9% oxygen by What is the empirical formula of1.56 the compound? 1.56
5 1.00 S
SO2 1.Let’s consider 100.0 g of compound, which contains 50.1 g of S and 49.9 g of O. We calculate the 3.12 number of moles of atoms of each. 5 2.00 O 1.56 50,1Exercise g 49,9 g You should now work 54. mol of S = mol of O = = 1,56 mol of S = 3,12 mol of O 32,1 g mol-1 16 g mol-1
2.We obtain a whole-number ratio between these numbers that gives the ratio of atoms in the empirical formula 1,56 = 1,00 S 1,56 2 3.12 The solution for Example 2-13 can be set up in tabular form. = 2,00 O 1,56
nt” mass of his n could of Moles mn wholems. But ame
SO
Element
Relative Mass of Element
S
50.1
O
49.9
Relative Number of Atoms (divide mass by AW)
Divide by Smallest Number
50.1 5 1.56 32.1 49.9 5 3.12 16.0
1.56 5 1.00 S 1.56 3.12 5 2.00 O 1.56
Smallest WholeNumber Ratio of Atoms
SO2
Derivation of Formulas from Composition
m mol x Fw
Sample exercise (Empirical Formula) g of an ionic compound is found to contain 6.072 g of Na, 8.474 g of S, and 6.336 g • 20.882 The solution for Example 2-13 can be set up in tabular form. of O. What is its empirical formula?
t” ass of s could Moles
hole. But me atio is
Element
Relative Mass of Element
Relative Number of Atoms (divide mass by AW)
Divide by Smallest Number
Smallest WholeNumber Ratio of Atoms
0,264 6,072 50.1g 1.56 = 0,264 mol = 1,00 1,00x2 = 2,00 5 1.56 5 1.00 S -1 1.56 0,264 23,0 32.1 g mol SO2 49.9 3.12 0,2645 2.00 O 8,474 g 5 3.12 49.9 g SO 8,474 1,00x2 = 2,00 = 0,264 mol = 1,00 16.0 -1 1.56 0,264 32,0 g mol 6,336 g 0,396 O 6,336 g 1,50x2 = 3,00as the =-1 0,396 way mol to solve simplest-formula This tabular format provides a convenient problems, = 1,50 16,0 g mol 0,264 next example illustrates. Na S
6,072 50.1 g
Empirical formula =
Na 2 S 2 O 3
ormula N.B. The ratio of atoms in the empirical formula must be a whole-number ratio (by definition). To convert
ratio 1:1:1.5 to ato wholenumber each number in and the 6.336 ratio was which gave the nicthe compound is found contain 6.072 gratio, of Na, 8.474 g of S, g of multiplied O. What isbyits2,simplest empirical formula Na2S2O3
Derivation of Formulas from Composition Practice exercise • The hormone norepinephrine is released in the human body during stress and increases the body’s metabolic rate. Like many biochemical compounds, norepinephrine is composed of carbon, hydrogen, oxygen, and nitrogen. The percent composition of this hormone is 56.8% C, 6.56% H, 28.4% O, and 8.28% N. What is the empirical formula of norepinephrine?
Derivation of Formulas from Composition Sample exercise (Percent composition) •
A 0.1647 g sample of hydrocarbon is burned in a C-H combustion train to produce 0.4931 g of CO2 and 0.2691 g of H2O. Determine the masses of C and H in the sample and the percentages of these elements in this hydrocarbon. 1. With a proportion, we use the observed masses to determine the masses of C and H in the original sample. There is one mole of carbons, (aw=12.01), in each mole of CO2, 44.01 g; there are two moles of hydrogen atoms, (aw=2.016), in each mole of H2O, 18.02 g:
mass of element mass of the compound
=
AW FW
g of C = 0.4931 g CO2 x12.01 g/mol C = 0.1346 g C 44.01 g/mol CO2 g of H = 0.2691 g H2O x 2.016 g/mol H = 0.0301 g H 18.02 g/mol H2O
2. Then we calculate the percentages by mass of each element:
% C = 0.1346 g C x 100% = 81.72% C 0.1647 g sample % H = 0.0301 g C x 100% = 18.28%H 0.1647 g sample Remember: Hydrocarbons are organic compounds composed entirely of hydrogen and carbon.
Molecular Formulas from Empirical Formulas To determine the molecular formula of a substance, both its empirical formula and its formula weight must be known. The molecular formula is a multiple of the empirical formula. E.g. Butane, C4H10. The empirical formula for butane is C2H5, but the molecular formula contains twice as many atoms; that is, 2x(C2H5)=C4H10. Benzene, C6H6. The empirical formula for benzene is CH. the molecular formula has six times as many atoms; 6x(CH)=C6H6.
The molecular formula for a compound is either the same as, or an integer multiple of, the (empirical) formula.
so
n=
molecular weight empirical formula weigh
A practical example: Combustion Analysis One technique for determining empirical formulas in the laboratory is combustion analysis, commonly used for organic compounds. When a compound containing carbon and hydrogen is combusted, the carbon is converted to CO2 and the hydrogen is converted to H2O. The amounts of CO2 and H2O produced are determined by measuring the mass increase in the CO2 SECTION 3.5 Empirical Formulas from Analyses 95 and H2O absorbers. From the masses of CO2 and H2O we can calculate the number of moles of C and H in the original sample and thereby the empirical formula.
mulas in the laboratory is ounds containing princi-
Sample combusted, producing CO2 and H2O
H2O and CO2 are trapped in separate absorbers
d hydrogen is completely Sample n in ! FIGURE 3.14, the O2 n is converted to H2O. H2O absorber CO2 absorber H2O produced are deterFurnace CO2 and H2O absorbers. ︎Apparatus for combustion analysis ate the number of moles Mass gained by each absorber corresponds to mass of the empirical formula. If CO2 or H2O produced If acan third is present in the compound, its mass can be determined by subtracting the measured mass beelement determined " FIGURE 3.14 Apparatus for masses C and H from the original sample mass. H from the of original samcombustion analysis.
Law of Multiple Proportions Two (or more) elements may form more than one compound (e.g. CO and CO2, NO and NO2, H2O and H2O2). The Law of Multiple Proportions states that
“When two elements, A and B, combine and form more than one compound, the ratio of the masses of element B, in each of the compounds, can be expressed by small whole numbers” E.g.: SO2 and SO3 provide an example. In SO2, two moles of oxygen combine with one mole of sulfur atoms In SO3, three moles of oxygen combine with one mole of sulfur atoms. Thus the ratio of oxygen atoms in the two compounds compared to a given number of sulfur atoms is 2:3. Many similar examples, such as CO and CO2 (1:2 oxygen ratio) and H2O and H2O2 (1:2 oxygen ratio), are known.
H2
O
2 amu
16 amu
1
:
8
H2O FW = 18
Law of Multiple Proportions Sample exercise (Empirical Formula) • What is the ratio of the numbers of oxygen atoms that are combined with a given number of nitrogen atoms in the compounds N2O3 and NO?
1.To compare the number of oxygen atoms, we must have equal numbers of nitrogen atoms. Because NO has half as many nitrogen atoms in its formula relative to N2O3, we must multiply it by a factor of 2 to compare the two elements on the basis of an equal number of nitrogen atoms. Once we show the number of atoms of each element present, we can cancel out the equal amounts of nitrogen atoms, leaving the ratio of oxygen atoms.
Oxygen ratio
=
N2O3 2N 3O = = 2 (NO) 2N 2O
3O 2O
=
3 2
The oxygen ratio in the two compounds N2O3 and NO is 3:2
Law of Multiple Proportions Practice exercise • Show that the compounds water, H2O, and hydrogen peroxide, H2O2, obey the Law of Multiple Proportions.
•
Nitric oxide, NO, is produced in internal combustion engines. When NO comes in contact with air, it is quickly converted into nitrogen dioxide, NO2, a very poisonous, corrosive gas. 1. What mass of O is combined with 3.00 g of N in (a) NO and (b) NO2? 2. Show that NO and NO2 obey the Law of Multiple Proportions.
Other Interpretations of Chemical Formulas Sample exercise (Empirical Formula) • Once we master the mole concept and the meaning of chemical formulas, we can use them in many ways. For example, what mass of chromium (Cr) is contained in 35.8 g of (NH4)2Cr2O7?
1.The formula tells us that each mole of (NH4)2Cr2O7 contains two moles of Cr atoms, so we first find the number of moles of (NH4)2Cr2O7. We know that the molar mass is 252.0 g/mol.
m mol x Fw
moles of (NH4)2Cr2O7
=
35.8 g = 252.0 g/mol
0.142 mols
2.Each mol of (NH4)2Cr2O7 contains 2 moles of Cr, so we convert the number of moles of (NH4)2Cr2O7 into the number of moles of Cr atoms it contains, using the proportion
moles of Cr
=
0.142 mol (NH4)2Cr2O7 x
2 mol Cr = 1 mol (NH4)2Cr2O7
0.284 mols
3.We then use the atomic weight of Cr to convert the number of moles of chromium atoms to the mass of chromium.
g of Cr
= 0.284 mol Cr x 51.99 g/mol Cr=
14.76 g Cr
Other Interpretations of Chemical Formulas Practice exercise • Mercury occurs as a sulfide ore called cinnabar, HgS. How many grams of mercury are contained in 578 g of pure HgS?
