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Chemical Formulas and Composition Stoichiometry

Chemical Formulas and Composition Stoichiometry. د. كريم محمد الصاوى استاذ مساعد الكيمياء الفيزيائية قسم الكيمياء - جامعة القصيم. Stoichiometry. A word derived from the Greek stoicheion, which means “first principle or element,” and metron, which means “measure.”.

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Chemical Formulas and Composition Stoichiometry

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  1. Chemical Formulasand CompositionStoichiometry د. كريم محمد الصاوى استاذ مساعد الكيمياء الفيزيائية قسم الكيمياء - جامعة القصيم

  2. Stoichiometry A word derived from the Greek stoicheion, which means “first principle or 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). The chemical formula for a substance shows its chemical composition. This represents the elements present as well as the ratio in which the atoms of the elements occur. The chemical formula for a single atom is the same as the symbol for the element. Thus, Na can represent a single sodium atom. A subscript following the symbol of an element indicates the number of atoms in a molecule. For instance, H2O is a molecule containing two hydrogen atoms and one oxygen atom. Number of hydrogen atoms

  3. Formula unit NaCl (NaCl)n A molecule is a definite group of atoms that are chemically bonded together Composition is described by the molecular formula E.g. H2O Formula unit: the lowest whole number ratio of ions present in an ionic compound Composition is described by the formula unit. E.g. CaCl2 The chemical formula for a single atom is the same as the symbol for the element. Subscript in the formula indicates the number of atoms

  4. In a chemical reaction atoms are re-arranged (Dalton’s Theory). for example, six H2 moleculesreact with three O2 molecules and rearrange to form six H2O molecules Equivalently, two H2 molecules react with one O2 molecule rearrange to form two H2O molecules So, the reaction takes place if we have the correct relativenumber of reactant PARTICLES (e.g. atoms, molecules,…). Atoms are so small to count (DIRECTLY), in the lab it is much easier to WEIGH chemicals. OK, If we get the correct relative weights of reactants, does this mean that we get the correct relative numbers of the reactant particles (atoms)? NO So. What!!!!!!!!!!!!!!!

  5. The Mole Chemists have adopted the mole concept as a convenient way to deal with the enormous numbers of molecules or ions in the samples they work with. • The mole is the SI unit for amount of substance, it isabbreviated mol. • The mole is defined as the amount of substance that contains as many particles (atoms, molecules, ..) as there are atoms in exactly 12 g of pure carbon-12 atoms. • The number of particles in one mole is known as Avogadro’s Number (NA) • Avogadro's number = 6.022x1023 = 602200000000000000000000.

  6. The mole is defined as the amount of substance that contains as many particles (atoms, molecules, ..) as there are atoms in exactly 12 g of pure carbon-12 atoms. One mole of hydrogen (H) contains the same number of particles as 12 g of carbon-12 One mole of helium (He) contains the same number of particles as 12 g of carbon-12 One mole of neon (Ne) contains the same number of particles as 12 g of carbon-12 The number of particle in one mole of hydrogen (H) = The number of particle in one mole of helium (He) = The number of particle in one mole of neon (Ne) = . . The number of particle in one mole of any substance = Avogadro's Number=6.022x1023

  7. 6.022x1023 carbon atoms = 12.000 g of carbon (one mole of carbon) The mole is defined as the amount of substance that contains as many particles (atoms, molecules, ..) as there are atoms in exactly 12 g of pure carbon-12 atoms. This means we now have Avogadro's number of carbon atoms (NA) = 6.022x1023 on the scales pan

  8. The mole is defined as the amount of substance that contains as many particles (atoms, molecules, ..) as there are atoms in exactly 12 g of pure carbon-12 atoms. Then, by definition: To get one mole of carbon all what we need to do is to weigh exactly 12 g of carbon. What if we need to get one mole of some other substance say Hydrogen, Helium, Sodium, …? How much do we need to weigh of these substance? (what is the molar mass of these substance?) It will be wrong to weigh 12 grams of each of these substances, 12 grams will not contain the same number of particles since atoms of different substances have different masses (Dalton’s Theory). Then, to proceed, we need to know how heavy each element is relative to carbon-12. In other words we need to know the relative atomic masses.

  9. DETERMINATION OF RELATIVE ATOMIC MASSES (WEIGHTS) Diagram of a simple mass spectrometer (top), showing the separation of neon isotopes (left). Neon gas enters an evacuated chamber, where neon atoms form positive ions when they collide with electrons. Positively charged neon atoms, Ne, are accelerated from this region by the negative grid and pass between the poles of a magnet. The beam of positively charged atoms is split into three beams by the magnetic field according to the mass-to-charge ratios. The three beams then travel to a detector at the end of the tube. ( The detector is shown here as a photographic plate)

  10. Thousands of experiments on the composition of compounds have resulted in the establishment of a scale of relative atomic masses (weights) based on the atomic mass unit (amu). • The atomic mass unit (amu), is defined as exactly 1/12of the mass of an atom of carbon-12. • Carbon-12 isotope was arbitrarily assigned a mass of exactly 12 atomic mass units • On this scale, the atomic weight (mass) of • hydrogen (H) is 1.00794 amu • Helium (He) is 4.0026 amu • sodium (Na) is 22.9897 amu, • magnesium (Mg) is 24.3050 amu. This tells us that Na atoms have nearly 23 times the mass of H atoms, and Mg atoms are about 24 times heavier than H atoms. Because of this relative scale, we can be sure that if we take 1:4 samples (by weight) of H and He, they will contain the same number of particles.

  11. The Molar Mass (M) • The molar mass of a substance is the mass of one mole of that substance. • Since one mole contains Avogadro's number of particles, then The Molar mass = the total mass of Avogadro's number of particles. • Since we have established a relative atomic mass scale relative to a common element (C-12), the following relationship is true. The mass of one mole (Molar Mass) of atoms of a pure element in grams is numerically equal to the atomic weight of that element in atomic mass units. • The molar mass units are grams/mole, also written as g/mol or g mol-1. • For instance, if you obtain a pure sample of the metallic element titanium (Ti), whose atomic weight is 47.88 amu, and measure out 47.88 g of it, you will have one mole, or 6.022 x1023 titanium atoms.

  12. One mole of atoms of some common elements. Back row (left to right): bromine, aluminium, mercury, copper. Front row (left to right): sulfur, zinc, iron. Each of these samples has a different mass but contains the same number of atoms; a very large number 6.022x1023 - Avogadro's number

  13. How many moles of atoms does 136.9 g of iron (Fe) metal contain? answer We are given the total mass of the Fe sample (m) = 136.9 g We look up the atomic mass of Fe = 55.85 amu Then The mass of one mole (molar mass M) of Fe= 55.85 g/mol Then

  14. How many atoms are contained in 2.451 mol of iron? answer We are asked about the total number of atoms (N) We are given the number of moles (n) =2.451 mol We know that one mole of atoms of an element contains Avogadro’s number (NA) of atoms, or 6.022 x1023 mol-1. Then

  15. Calculate the average mass of one iron atom in grams. answer The atomic mass of Fe = 55.85 amu Then The mass of one mole (molar mass M) of Fe= 55.85 g mol-1 This is the mass of Avogadro’s number of atoms Then Thus, the average mass of one Fe atom is only 9.274 x10-23 g, that is, 0.00000000000000000000009274 g

  16. FORMULA WEIGHTS, MOLECULAR WEIGHTS AND MOLES So, we’ve learnt how to get the molar mass of atomic substances using the relative atomic mass scale based on C-12 How about the molar mass of molecules and ionic substances? The mass of one mole (molar mass) of a substance in grams is numerically equal to the formula weight of that substance in atomic mass units. The formula weight = sum of the atomic masses (weights) of the elements in the formula

  17. Calculate the formula weight (mass) and molar mass of each of the following a) chloroform, CHCl3; b). iron(III) sulfate, Fe2(SO4)3; c) water, H2O Answer a) chloroform, CHCl3 Then the molar mass of CHCl3= 119.4 g mol-1 b) iron(III) sulfate, Fe2(SO4)3 Then the molar mass of Fe2(SO4)3= 399.9 g mol-1 c) water, H2O

  18. Molecule, Mole and Molar Mass Molar mass of H2O = mass of one mole of H2O = Total mass of 6.022x1023 molecules of H2O = Formula weight in grams of H2O = 18 g mol-1 One molecule of H2O One mole of H2O (6.022x1023 molecules of H2O)

  19. How many moles of substance are contained in each of the following samples? (a) 18.3 g of NH3 (b) 5.32 g of ammonium bromide NH4Br (c) 6.6 g of PCl5 (d) 215 g of Sn answer Br 79.9 N 14.0 P 30.9 Cl 35.5 C 12.0 S 32.0 Sn 118.7

  20. What mass, in grams, should be weighed for an experiment that requires 1.54 mol of (NH4)2HPO4? answer

  21. How many (a) moles of O2, (b) molecules of O2 , and (c) O atoms are contained in 40.0 g of oxygen gas (dioxygen)

  22. Calculate the number of hydrogen atoms in 39.6 g of ammonium sulfate, (NH4)2SO4. One moleculeof(NH4)2SO4 contains 8 hydrogen atoms One mole of (NH4)2SO4contains 8 x NA hydrogen atoms n mole of (NH4)2SO4contains 8 x NA x n hydrogen atoms

  23. How many hydrogen atoms are contained in 125 grams of butane, C4H10 Answer One moleculeofC4H10 contains 10 hydrogen atoms One mole of C4H10 contains 10 x NA hydrogen atoms n mole of C4H10 contains 10 x NA x n hydrogen atoms

  24. What mass of chromium is contained in 35.8 g of (NH4)2Cr2O7? answer Reading the given chemical formula tells us that One mole of(NH4)2Cr2O7 contains 2 moles of Cr We can succinctly write this in terms of molar masses as: One molar mass of (NH4)2Cr2O7 → 2 x Molar mass of Cr 35.8 g of (NH4)2Cr2O7 → X

  25. What is the formula…? When a chemist has discovered a new compound, the first question to answer is, What is the formula? To answer, you begin by analyzing the compound to determine amounts of the elements for a given amount of compound. This is conveniently expressed as percentage composition—that is, as the mass percentages of each element in the compound. But, what do we really mean by a chemical formula?

  26. The simplest, or empirical, formula for a compound is the smallest whole-number ratio of atoms present. The molecular formula indicates the actual numbers of atoms present in a molecule of the compound. It may be the same as the simplest formula or else some whole-number multiple of it. E.g. The molecular formula for hydrogen peroxide is H2O2 This tells us that there are two hydrogen atoms and two oxygen atoms in the molecule The empirical formula for hydrogen peroxide is HO This tells us that the ratio between the hydrogen and oxygen atoms in the molecule is 1 : 1 How to get these formula?

  27. PERCENT COMPOSITION AND FORMULAS OF COMPOUNDS Suppose that A is a part of something—that is, part of a whole. It could be an element in a compound or one substance in a mixture. We define the mass percentage of A as the parts of A per hundred parts of the total, by mass. That is If the formula of a compound is known, its chemical composition can be expressed as the mass percent of each element in the compound (percent composition).

  28. E.g. Calculate the percent composition by mass of HNO3. Answer We are not given the mass of HNO3 but we can calculate its molar mass and then derive its percent composition from it. Nitric acid is 1.6% H, 22.2% N, and 76.2% O by mass. N.B. All samples of pure HNO3 should have this composition, according to the Law of Definite Proportions.

  29. DERIVATION OF FORMULAS FROM ELEMENTAL COMPOSITION Get the relative number of atoms for each element ( mass/gram atomic mass AW) Chemical formula corresponds to the smallest whole number ratio of atoms Divide by the smallest number Convert to whole numbers mass of each element To get a chemical formula for a compound we need the number (or the relative number) of atoms in the compound. So if we have the mass of each element we can proceed as follows: A 20.882-gram sample of an ionic compound is found to contain 6.072 grams of Na, 8.474 grams of S, and 6.336 grams of O. What is its empirical ( simplest) formula? answer

  30. A 0.1014 g sample of purified glucose was burned in a C-H combustion train to produce 0.1486 g of CO2 and 0.0609 g of H2O. An elemental analysis showed that glucose contains only carbon, hydrogen, and oxygen. Determine the masses of C, H, and O in the sample. Answer 1 mole of CO2 →1 mole of C Molar mass of CO2= 12.0 g mol-1 + 2x16.0 g mol-1 = 44.0 g mol-1 Molar mass of C = 12.0 g mol-1 44 g mol-1 of CO2 → 12.0 g mol-1 of C 0.1486 g of CO2 → ? g C 1 mole of H2O→2 mole of H Molar mass of H2O= 2x1.0 g mol-1 +16.0 g mol-1 = 18.0 g mol-1 Molar mass of H = 1.0 g mol-1 18 g mol-1 of H2O → 2.0 g mol-1 of H 0.0609 g of H2O → ? g H

  31. Get the relative number of atoms for each element ( mass/gram atomic mass AM) Chemical formula corresponds to the smallest whole number ratio of atoms Divide by the smallest number Convert to whole numbers mass of each element In the previous example a 0.1014 -gram sample of glucose was found to contain 0.04055 grams of C, 0.00681 grams of H, and 0.0540 grams of O. What is the empirical ( simplest) formula of glucose? answer

  32. Molecular formula = k x simplest formula k = integer 1,2,3,…. DETERMINATION OF MOLECULAR FORMULAS As we have seen percent composition data yield only simplest (empirical) formulas. To determine the molecular formula for a molecular compound, both its simplest formula and its molecular weight must be known. If k=1, then the molecular formula and empirical formula are the same E,g. H2O

  33. In the previous example, the simplest formula for glucose was found to be CH2O. Other experiments have shown that its molecular weight is approximately 180 amu. Determine the molecular formula of glucose. Answer

  34. Deduce the empirical formulas of the following molecular formulas • C6H12O6 b) H2O2 b) H2O • answer We start from the definition: Molecular formula = k x simplest formula k = integer 1,2,3,…. Then we need to find k …k will be the largest common whole number divisor of the numbers of atoms in the molecular formula • C6H12O6 • The molecular formula tells us that for each C6H12O6 molecule there are: • 6 C atoms • 12 H atoms • 6 O atoms • Common divisors of 6,12 and 6 are 1,2,3 and 6 • We choose the largest, k=6 • The empirical formula is then C6/6 H12/6 O6/6 = CH2O

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