Chemical Formulas Methane CH 4 MOLECULAR FORMULAS representation

Chemical Formulas Methane CH 4

MOLECULAR FORMULAS: representation of molecules in terms of their constituent atoms -– H 2, H 2 O DIATOMIC - two atoms; example: H 2, N 2, CO (carbon monoxide) TRIATOMIC - three atoms; example: CO 2 (carbon dioxide), O 3(ozone) POLYATOMIC - more than three atoms; examples: NH 3 (ammonia), CH 4 (methane), C 2 H 6 O 2 (ethylene glycol)

What limits the “boundary” of a molecule? Atoms in molecules are held together by strong interactions called CHEMICAL BONDS Interactions between neutral atoms in a molecule (e. g. H 2, H 2 O) is called a COVALENT bond, forming covalent compounds Interactions between charged elements (IONS) result in a different kind of chemical bond, called an IONIC BOND Compounds formed via interactions between ionic (charged) elements are called ionic compounds.

In a crystal of Na. Cl a Na+ ion interacts with the neighboring Cl- ion, hence a Na. Cl “molecule” does not exist. However, if one counted the number of sodium ions and the number of Cl- ions in a given volume we would find that the ratio of Na+ : Cl- is 1: 1. A crystal of sodium chloride is electrically neutral and its CHEMICAL FORMULA can be expressed as Na. Cl.

Chemical formulas indicate the constituent elements in a compound (covalent or ionic) The term molecular formula refers specifically to covalent compounds. How are chemical formulas determined?

ELEMENTAL ANALYSIS Determination of the relative amounts of the elements in a compound Determines the EMPIRICAL FORMULA which is the simplest possible formula and indicates the relative amounts of constituent elements For example, the molecular formula of hydrogen peroxide is H 2 O 2; its empirical formula is HO. The empirical and the chemical formula can be the same, for example, H 2 O. For ionic compounds, the empirical formula is the same as the chemical formula (Na. Cl)

Problem An oxide of nitrogen is analyzed and found to contain 25. 9% N and 74. 1% O. What is the empirical formula of the compound? In 100. 0 g of compound: # moles of N = 25. 9 g N / (14. 01 g/mol) = 1. 85 mol # moles of O = 74. 1 g O / (16. 00 g/mol) = 4. 63 mol Ratio of O: N : : 2. 5: 1 Must be whole numbers – N 2 O 5

Combustion Analysis - elemental analysis of organic compounds. Organic compounds contain predominantly C and H. Minor elements include O, N, S, Cl, P, etc. In combustion analysis the organic compound is burnt in the presence of oxygen (O 2) to form carbon dioxide (CO 2), water (H 2 O) and other gases (nitrogen (N 2), sulfur dioxide (SO 2))

A sample of a compound that is known to contain only carbon, hydrogen, and oxygen is combusted, and the CO 2 and H 2 O produced are trapped and weighed. The original sample weighed 8. 38 g and yielded 16. 0 g CO 2 and 9. 5 g H 2 O. What is the empirical formula?

Need to determine the number of moles of C and H in the compound, and then determine the mass of O in the compound. Moles of C = (16. 0 g CO 2 / 44. 0 g/mol CO 2) x (1 mol C/1 mol CO 2) = 0. 364 mol C Moles of H = (9. 80 g H 2 O / 18. 0 g/mol H 2 O) x (2 mol H / 1 mol H 2 O) = 1. 09 mol H

Mass of C in compound = 0. 364 mol C x 12. 0 g/mol C = 4. 37 g C Mass of H in compound = 1. 09 mol H x 1. 01 g/ mol H = 1. 10 g H Mass of O in compound = 8. 38 g - 4. 37 g - 1. 10 g = 2. 91 g O Moles of O in compound = (2. 91 g O / 16. 0 g/mol O) = 0. 182 mol O Mole ratio of C : H : O is 0. 364 : 1. 09 : 0. 182 or 2: 6: 1 Hence empirical formula is C 2 H 6 O

Determining Chemical Formulas The empirical formula tells you the simplest ratio of the individual elements in the compound. For an ionic compound this information is enough. For a molecular compound this may not be enough since the empirical formula may not be the molecular formula. Knowledge of the MOLAR MASS of the compound and its empirical formula, allows the molecular formula to be determined.

Elemental analysis of a sugar shows that it consists of 40. 0% carbon (C), 6. 7% hydrogen (H), 53. 3% oxygen (O). The molar mass of the compound was found to be 180. 0 g/mol. What is the molecular formula of the compound?

moles of C in 100. 0 g of compound = (mass of C g)/(atomic mass g/mol) = 40. 0/12. 01 = 3. 33 mol C in 100. 0 g of compound moles of H in 100. 0 g of compound = (mass of H g)/(atomic mass g/mol) = 6. 7/1. 01 = 6. 7 mol H in 100. 0 g of compound moles of O in 100. 0 g of compound = (mass of O g)/(atomic mass g/mol) = 53. 3/16. 00 = 3. 33 mol O in 100. 0 g of compound Ratio of C: H: O : : 1: 2: 1; hence empirical formula is CH 2 O

Molar mass of empirical formula = 12. 01 + 2(1. 01) + 16. 00 = 30. 0 g/mol Ratio of molar mass of compound : molar mass of empirical formula 180. 0/30. 0 = 6. 0 molecular formula is (CH 2 O) 6 or C 6 H 12 O 6

Chemical Reactions & Equations In a chemical reaction elements and/or compounds collide, interact and react to form new compounds. The compounds that come together are called the REACTANTS and the new compounds formed are the PRODUCTS.

An important aspect of a chemical reaction is that MASS IS ALWAYS CONSERVED - i. e. the total mass of the reactants must equal the total mass of the products. To ensure that mass is conserved, we have to keep track of the number of atoms of each element in the reactants and number of atoms of each element in the products

Writing Chemical Equations H 2 + O 2 H 2 O reactants products Mass has not been conserved - equation is not BALANCED Need to make sure that the number of atoms of a given element is the same on either side.

H 2 + O 2 H 2 O H is balanced, but O is not. To balance O, multiply H 2 O by 2 H 2 + O 2 2 H 2 O Now O is balanced but not H (4 H’s on right, 2 on left) Multiply H 2 by 2 2 H 2 + O 2 2 H 2 O

2 H 2 + O 2 2 H 2 O Now the equation is balanced. However, for a complete chemical equation the state of all reactants and products must be included 2 H 2(g) + O 2 (g) 2 H 2 O (g) g - gas; l - liquid; s- solid aq - denotes a solution of a solute in water (solvent) Note: the subscripts in the molecular formula must never be changed in balancing equations. Changing the subscript corresponds to a different molecular formula, hence a different molecule, with completely different properties

“Rules” to balance chemical equations Consider the reaction of methane, CH 4, burning in air to produce CO 2 and H 2 O (combustion reaction) CH 4 + O 2 CO 2 + H 2 O It is usually best to first balance those elements that occur in the fewest chemical formulas on each side of the equation. So in this example balance C and H first

C is already balanced. However 4 H’s on the left and 2 on the right Hence, CH 4 + O 2 CO 2 + 2 H 2 O Now balance O CH 4 + 2 O 2 CO 2 + 2 H 2 O Now specify the state of each reactant and product CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g)

Ag 2 S(s) + KCN(aq) + O 2(g) + H 2 O(l) KAg(CN)2(aq) + S(s) + KOH(aq) Sometimes it is hard to balance a chemical reaction by “inspection” Can the use algebraic expressions to balance equations

a Ag 2 S(s) + b KCN(aq) + c O 2(g) + d H 2 O(l) 1 KAg(CN)2(aq) + e S(s) + f KOH(aq) Element Reactant Product Ag 2 a 1 S a e K b 1+f C b 2 N b 2 O 2 c + d f H 2 d f

2 a = 1 ; a = 0. 5 b=2 a = e ; e = 0. 5 b = 1 + f; f = 1 2 d = f; d = 0. 5 2 c + d = f; c = 0. 25 Divide all by 0. 25 a = 2; b = 8; c = 1; d = 2; e = 2; f = 4 Hence 2 Ag 2 S(s) + 8 KCN(aq) + O 2(g) + 2 H 2 O(l) 4 KAg(CN)2(aq) + 2 S(s) + 4 KOH(aq)

Chemical Stoichiometry Reactants are consumed and products are formed in definite proportions. These proportions are given by the coefficients in the balanced equations for chemical reactions The calculation of the quantities of reactants and products is called STOICHIOMETRY. Stoichiometry is the use of chemical equations to calculate quantities of substances that take part in chemical reactions.

To do a stoichiometric calculation, the chemical equation for the calculation must be balanced. Equations are read in terms of moles of reactants and product 2 H 2(g) + O 2 (g) 2 H 2 O(g) 2 moles of H 2 (g) reacts with 1 mole of O 2 (g) to form 2 moles of H 2 O(g) Or: 2 No molecules of H 2 (g) + No molecules of O 2 (g) 2 No molecules of H 2 O(g)

Problem Consider the reaction of 100 g of H 2(g) with sufficient O 2 (g) to produce the stoichiometric quantity of H 2 O(g). (stoichiometric quantities are the exact amounts of reactants and products predicted by balanced equations). Calculate the mass of H 2 O formed. Need a balanced equation for this reaction 2 H 2(g) + O 2 (g) 2 H 2 O(g) To find the mass of water formed, need to find the number of moles of H 2 that reacted: (mass of H 2) (molar mass of H 2) = (100. g H 2)/(2. 02 g/mol) = 49. 5 mol H 2

2 moles of H 2 reacts with 1 mole of O 2 to form 2 moles of H 2 O => 49. 5 moles of H 2 will form 49. 5 moles of H 2 O Hence, the mass of H 2 O(g) formed =(49. 5 moles H 2 O) x (18. 02 g/mol) = 892 g H 2 O

Problem: Diethyl ether (C 4 H 10 O) combusts in air, reacting with O 2 to form H 2 O and CO 2. How many grams of CO 2 would be produced if 350. 0 m. L of diethyl ether were combusted in an unlimited amount of oxygen? The density of diethyl ether is 0. 713 g/m. L. First write a balanced equation ? C 4 H 10 O(l) + ? O 2(g) ? H 2 O(l) + ? CO 2(g) Then convert the volume of diethyl ether to mass Next, determine the number of moles of diethyl ether reacted Determine the number of moles of CO 2 formed based on the stoichiometry of the reaction Finally determine the mass of CO 2 produced Answer: 592 g CO 2

For reactions occurring in solution, instead of using moles as a unit of quantity, define moles/liter = MOLARITY Molarity is defined as the number of moles of a solute per liter of solution. Note: the solute can be a solid, liquid, or gas dissolved in a solvent.

Problem For the reaction: 2 Pb. O 2(s) + 4 HNO 3(aq) 2 Pb. NO 3(aq) + 2 H 2 O (l) + O 2(g) What volume of 7. 91 M solution of nitric acid, HNO 3, is just sufficient to react with 15. 9 g of lead dioxide, Pb. O 2? Answer: 0. 0168 L

Volume Relationship of Gases in Chemical Reactions Gay Lussac’s law of combining volumes – states that the volumes of gaseous reactants and products stand in ratios of simple integers, as long as those volumes are measured at the SAME temperature and pressure. These integers are the same as the integers used to balance the chemical equation. A balanced chemical equation therefore provides a relationship between the volumes of gases reacting.

Problem: Gaseous H 2 and N 2 react according to the equation 3 H 2 (g) + N 2 (g) 2 NH 3(g) If in a given reaction 5. 00 L of NH 3 are formed, what volume of H 2 reacted, assuming that the temperature and pressure before and after the reaction are the same. From the balanced reaction above, assuming the same temperature and pressure before and after the reaction, 2 volumes of NH 3 are formed from 3 volumes of H 2 Hence, 5. 00 L of NH 3 are formed from (3/2)x 5. 00 L = 7. 5 L of H 2

Limiting Reactants and Product Yields If everything went perfectly in the reaction : Fe(s) + S (s) Fe. S (s) 55. 8 g (1 mole) of iron will react with 32. 1 g (1 mole) of S to form 87. 9 g (1 mole) of Fe. S. If stoichiometric amounts of reactants were used, and assuming that there are no competing factors to limit the amount of products being formed, then a stoichiometric amount of product is formed

If we were to mix reactants together in non-stoichiometric amounts then what determines the amount of product formed? The reactant which is first used up determines the amount of product formed. This reactants is called the LIMITING REACTANT. The other reactants are in EXCESS when the reaction stops. If additional amounts of the limiting reactant is added the reaction starts again.

Problem: Calcium carbonate Ca. CO 3(s) is decomposed by HCl(aq) to give Ca. Cl 2 (aq), CO 2(g) and H 2 O(l). If 10. 0 g of Ca. CO 3 are treated with 10. 0 g of HCl, how many grams of CO 2 are generated? First write a balanced equation for the reaction: Ca. CO 3 (s) + 2 HCl(aq) Ca. Cl 2(aq) + H 2 O(l) + CO 2(g) 1 mole of Ca. CO 3 reacts with 2 moles of HCl to form 1 mole of CO 2 Moles of Ca. CO 3 = (10. 0 g)/(100. 0 g/mol) = 0. 100 mole Ca. CO 3 Moles of HCl = (10. 0 g)/(36. 5 g/mol) = 0. 274 mol HCl

Hence 0. 1 mole of Ca. CO 3 reacts with 0. 2 moles of HCl. Since the amount of HCl present is greater that 0. 2, HCl is in excess and Ca. CO 3 is the limiting reactant. Hence, 0. 1 mole of CO 2 formed. Mass of CO 2 formed = 0. 1 mol x 44. 01 g/mol = 4. 40 g CO 2

Product Yields In the previous calculation 4. 40 g is the amount of CO 2 we would expect to be formed 4. 40 g CO 2 is the CALCULATED or the THEORETICAL PRODUCT YIELD. This assumes that the reaction goes to completion, and that there are no competing factors that may reduce the amount of CO 2 formed. The measured amount of product formed is called the ACTUAL YIELD which is often smaller than theoretical yield.

Percentage Yield = (actual yield) x 100 % (theoretical yield) The larger the % yield the more cost effective is the process and hence a more likely candidate for industrial scale processes (other factors are also important: the nature of the byproducts - are they environmentally safe -, the cost of the starting materials, etc. )

Problem: The principle step in the recovery of elemental iron (Fe) from iron oxide (Fe. O) is a process known as reduction. In this reaction hydrogen (H 2) at elevated temperatures is used as the reducing agent. Calculate the percent yield of Fe if 1. 00 g of H 2 reacts with 30. 0 g of Fe. O and produces 19. 5 g of Fe. The reaction between Fe. O and H 2 is: Fe. O(s) + H 2 (g) Fe(s) + H 2 O(g) First we need to determine the limiting reactant Moles of H 2 in reaction mixture = 1. 00 g / (2. 02 g/mol) = 0. 500 mol Moles of Fe. O reaction mixture = 30. 0 g / (71. 85 g/mol) = 0. 418 mol

From the stoichiometry of the equation Fe. O and H 2 react in a 1: 1 ratio. Hence, Fe. O is the limiting reactant The number of moles of Fe that should be formed = 0. 418 moles Mass of Fe that should be formed = 0. 418 mole x 55. 85 g. mol = 23. 3 g of Fe. The actual yield of Fe. O is 19. 5 g % yield = (19. 5)x 100 = 83. 7% 23. 3

Ideally for an industrial process, the % yield should be large. Performing stoichiometric calculations, determining % yields, are important in analyzing the “success” of a chemical reaction. Sometime, a reaction performed on a lab-scale may have good product yields but when the reaction is scaled up to an industrial process, the yield may be lower.