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# Maths and chemistry chapter

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### Maths and chemistry chapter

1. 1. G I M N A Z J U M I M . A N N Y W A Z Ó W N Y , G O L U B - D O B R Z Y Ń MATHS AND CHEMISTY
2. 2. Page 2 THIS EBOOK WAS PREPARED AS A PART OF THE COMENIUS PROJECT WWWHHHYYY MMMAAATTTHHHSSS??? by the students and the teachers from: GIMNAZJUM IM. ANNY WAZÓWNY IN GOLUB-DOBRZYŃ IN POLAND This project has been funded with support from the European Commission. This publication reflects the views only of the author, and the Commission cannot be held responsible for any use which may be made of the information contained therein.
3. 3. Page 3 II.. BBAALLAANNCCIINNGG CCHHEEMMIICCAALL EEQQUUAATTIIOONNSS IIII.. HHOOWW TTOO CCAALLCCUULLAATTEE DDEENNSSIITTYY IIIIII.. PPEERRCCEENNTTAAGGEE CCOOMMPPOOSSIITTIIOONN IIVV.. DDEETTEERRMMIINNIINNGG EEMMPPIIRRIICCAALL AANNDD MMOOLLEECCUULLAARR FFOORRMMUULLAASS VV.. CCOONNCCEENNTTRRAATTIIOONN BBYY PPEERRCCEENNTT
4. 4. Page 4 Picture from: www.wyckoffps.org Chemists use math for a variety of tasks. They balance the equation of a chemical reaction, use mathematical calculations that are absolutely necessary to explore important concepts in chemistry, and utilize dimensional analysis to find any range of information about reactions from finding the mass of chemicals reacted to the concentration of a chemical in a solution. Math is also used to calculate energy in reactions, compression of a gas, grams needed to add to a solution to reach desired concentration, and quantities of reactants needed to reach a desired product. It is important to know how to mathematically handle chemistry problems in order to understand what they mean and how to prepare specific quantities of chemicals. Balancing chemical equations Stoichiometry - is a branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. Stoichiometry is the mathematics behind the science of chemistry. A chemical equation is an easy way to represent a chemical reaction—it shows which elements react together and what the resulting products will be. By the Law of Conservation of Mass, the number of atoms must be the same on both sides because these atoms cannot be created or destroyed in a reaction. The number of atoms that we start with at the beginning of the reaction must equal the number of atoms that you end up with. When the number of atoms of reactants matches the number of atoms of products, then the chemical equation is said to be balanced. We would like to present a simple method of defining the coefficients in the equations of chemical reactions with the help of a system of linear algebraic equations that describes the material balance in a chemical reaction. Example 1: CH4 + O2 → CO2 + H2O I. First we use each element to produce an equation involving the coefficient letters. We need to find the smallest possible positive integers a, b, c, and d such that the following chemical equation.
5. 5. Page 5 Picture from: www.earthtimes.org aCH4 + bO2 → cCO2 + dH2O is balanced. Carbon: a = c There is 1 carbon atom in the a term and 1 in the c term. Hydrogen: 4a = 2d There are 4 hydrogen atoms in the a term and 2 in the d term. Oxygen: 2b = 2c + d There are 2 oxygen atoms in the b term and 2 in the c term and 1 in the d term. II. As a result we get a system of three linear equations with four unknowns: III. This system has an infinite number of solutions, but we have to get the minimal natural values only. Since we would like the smallest integral solutions, = 1 works well. The system has the following solution - the coefficients are = 1, = 2, = 1 and = 2 When writing our balanced equation, the coefficient 1 is assumed and can be omitted, yielding the formula: CH4 + 2O2 → CO2 + 2H2O The balanced chemical equation has one mole of methane reacting with two moles of oxygen gas to form one mole of carbon dioxide and two moles of water. Example 2: Plants use the process of photosynthesis to convert carbon dioxide and water into glucose and oxygen. This process helps remove carbon dioxide from the atmosphere. Balance the following equation for the production of glucose and oxygen from carbon dioxide and water. CO2 + H2O → C6H12O6 + O2 This equation needs to be balanced. We must find coefficients a, b, c and d in the reactants and products - rewrite the equation as: aCO2 + bH2O → cC6H12O6 + dO2 where the numbers of carbon, hydrogen, and oxygen atoms are the same on both sides of the equation: Carbon: a = 6c Oxygen: 2a + b = 6c + 2d Hydrogen: 2b = 12c
6. 6. Page 6 Picture from: www.commons.wikimedia.org Since we would like the smallest integral solutions, = 6 works well and the coefficients are = 6, = 6, = 1 and = 6 Examples of Balancing Chemical Equations Consider the unbalanced reactions. 1. FeCl3 + NH4OH → Fe(OH)3 + NH4Cl Assign each molecule a variable a, b, c, d since we have 4 expressions in the reaction We need to find the smallest possible positive integers a, b, c, and d such that the following chemical equation. aFeCl3 + bNH4OH → cFe(OH)3 + dNH4Cl We can obtain a set of linear equations in these variables by considering the number of times each type of atom occurs on each side of this equation 1) Fe: a = c There is 1 iron atom in the a term and 1 in the c term. Cl: 3a = d There are 3 chlorine atoms in the a term and 1 in the d term N: b = d There is 1 nitrogen atom in the b term and d in the c term H: 5b = 3c + 4d There are 5 hydrogen atoms in the b term and 3 in the c term and 4 in d term. O: b = 3c There is 1 oxygen atom in the a term and 1 in the c term 2) As a result we will get a system of five linear equations with four unknowns: 3) 4) FeCl3 + 3NH4OH → Fe(OH)3 + 3NH4Cl
7. 7. Page 7 2. aNaCl + bSO2 + cH2O + dO2 → eNa2SO4 + f HCl From this equation we obtain the following relations in the unknowns a, b, c, d, e, and f: 1) Na: a = 2e Cl: a = f S: b = e O: 2b + c + 2d = 4e H: 2c = f 2) 3) 4) To calculate the smallest possible positive integer value of a, we have to find the least common denominator of b, c, d, and e which in this case is 4. If we work out the above equations we calculate the values of our 5 variables to be: For a = 4 we have: b = 2 c = 2 d = 1 e = 2 f = 4 4NaCl + 2SO2 + 2H20 + 02 → 2Na2SO4 + 4 HCl
8. 8. Page 8 Picture from: www.chemistrysmostwanted.wikispaces.com 3. aCaCO3 + bHNO3 → cCa(NO3)2 + dH2O + eCO2 Specifying the values a, b, c, d and e for the coefficients of this equation we have: 1) Ca: a = c C: a = e H: b = 2d = 2a N: b = 2c = 2a O: 3a + 3b = 6c + d + 2e Solving simultaneously and using the smallest integers we have 1) 2) For a = 1 b = 2a = 2 c = a = 1 d = a = 1 e = a = 1 CaCO3 + 2 HNO3 → Ca(NO3)2 + H2O + CO2 How is marble eroded by acid rain? Atmospheric sulfur dioxide combines with rainwater to create sulfurous acid. The primary component of marble is calcium carbonate (CaCO3). The sulfurous acid reacts with the CaCO3 in the marble and dissolves it. Marble statues (CaCO3) attacked by acid rain (containing HNO3). 4. aHCl + bK2CO3 → cCO2 + dH2O + eKCl The equation for each atom looks like: 1) H: a = 2d d = Cl: a = e K: 2b = e b = C: b = c c =
9. 9. Page 9 O: 3b = 2c + d So we have now after some canceling: 2) b = c = d = e = For a = 2 we have: 3) b = 1 c = 1 d = 1 e = 2 2HCl + K2CO3 → CO2 + H2O + 2KCl
10. 10. Page 10 How to calculate density Density is a characteristic property of a substance. The density of a substance is the relationship between the mass of the substance and how much space it takes up (volume). The density, or more precisely, the volumetric mass density, of a substance is its mass per unit volume. Mathematically, density is defined as mass divided by volume : v m volume mass d  To calculate the specific gravity (S.G.) of an object, you compare the object's density to the density of water. Examples of densities: Solids: Liquids: Gases: Silver = 10.49 Milk = 1.020-1.050 Air = 0.001293 Aluminum = 2.7 Glucose = 1.350-1.440 Argon = 0.001784 Diamond = 3.01-3.25 Glycerine = 1.259 Chlorine= 0.0032 Gold = 19.3 Flourine = 0.001696 Magnesium = 1.7 Helium = 0.000178 Platinum = 21.4 Neon = 0.0008999 Example 1 Calculate the density of the cube made of silver, whose weight is 262.5 grams and the volume is 25 cm ³. Given: Calculate m = 262.5 g d = ? V = 25 cm³ 1 d = Answer: Density of silver is 10.5 . Example 2 There is 250 cm ³ of ethyl alcohol in the glass vessel. The volume of alcohol is 0.197 kg Calculate the density of alcohol and enter the result in grams per cubic centimeter. Given: Calculate: m = 0.197 kg d = ? V = 250 cm³ 1 m = 0.197kg= 0.197 g
11. 11. Page 11 Picture from: www.bonnibrodnick.com 2 d = The density of ethyl alcohol is 0.79 Example 3 Calculate the mass of 300 cm³ gasoline which density is 0.75 . Given: Calculate: V = 300 cm³ m = ? d = 0.75 1 d = m = d V 2 m = 0.75 300cm³ = 225 g Answer: The mass of gasoline is 225g. Example 4 A container of volume 0.05m3 is full of ice. When the ice melts into water, how many kg of water should be added to fill it up? (density of ice = 900 ; density of water = 1000 ) Given: Calculate: dice = 900 mwater = ? dwater = 1000 Vice = 0.05 m³ 1 mice = d V mice = 2 We should add 5 kg of water to fill the container up. Example 5 A rubber ball has a radius of 2.5 cm. The density of rubber is 1.2 . What is the mass of the ball? Given: Calculate: r = 2.5 cm m = ? d = 1.2 1 First we calculate the volume of a ball: 2 Answer: The mass of the ball is about 78.5 g.
12. 12. Page 12 Example 6 A 5.6-gram marble put in a graduated cylinder raises the water from 30 mL to 32 mL. What is the marble’s density? Given: Calculate: m = 6g d = ? 1 First we calculate the volume of marble: 2 d = The density of marble is . Example 7 A small rectangular slab of lithium has the dimensions 20.9 mm by 11.1 mm by 11.9 mm. Its mass is 1.49·103 mg. What is the density of lithium in ? Given: Calculate: a = 20.9 mm = 2.09 cm d = ? b = 11.1 mm = 1.11 cm c = 11.9 mm = 1.19 cm m = 1.49·103 mg = 1490 mg = 1.49 g 1 V 2 d = The density of lithium is 0.53 . Example 8 Find the mass of air inside a room measuring 10m×8m×3m, if the density of air is 1.28 . Given: Calculate: a = 10m m = ? b = 8m c = 3m d = 1.28 1 V 2 m = d V m = The mass of the air inside the room is 2mL
13. 13. Page 13 Picture from:www.monarchbearing.com Example 9 You have two stainless steel balls. The larger has a mass of 25 grams and a volume of 3.2cm3. The smaller has a mass of 10 grams. Calculate the volume of the smaller ball. Given: Calculate: m1 = 25g V2 = ? V1 = 3.2cm3 m2 = 10g 1 dsteel = 2 V2 = = The volume of the smaller ball is Since density is a characteristic property of a substance, each liquid has its own characteristic density. The density of a liquid determines whether it will float on or sink in another liquid. A liquid will float if it is less dense than the liquid it is placed in. A liquid will sink if it is more dense than the liquid it is placed in. Example 10 A rectangular object is 10 centimeters long, 5 centimeters high, and 20 centimeters wide. Its mass is 800 grams. Will the object float or sink in water? Remember that the density of water is about 1 . Given: Calculate: a = 10cm d = ? b = 5cm c = 20cm m = 800g 1 V 2 d = The object will float.
14. 14. Page 14 carbon oxygen carbon hydrogen oxygen Picture from: www.annekeckler.com Percentage composition Percentage composition is just a way to describe what proportions of the different elements there are in a compound. If you have the formula of a compound, you should be able to work out the percentage by mass of an element in it. %Composition A= Example 1 What is the percentage composition of carbon and oxygen in ? First we need to find the mass of the compound. Molar mass of compound: 12.01+ 2 Next we need to find the mass of carbon and oxygen in the compound. Molar mass of carbon: 12.01 Molar mass of oxygen: 32 Then we should divide the mass of each element by the mass of the compound and multiply by 100%. The percentage composition of carbon is: %C = The percentage composition of oxygen is: % O = 72.71% Example 2 What is the percentage composition by mass of the elements in the compound We start by finding the atomic weights. Molar mass of C: = 12.01 Molar mass of H: = 1.01 Molar mass of O: = 16.00 Work out the molecular weight of glucose: 6 12.01 + 12 1.01 + 6 16 = 180
15. 15. Page 15 potassium chromium oxygen Picture from: www.commons.wikimedia.org Picture from: www.aromaticscience.com Mass of carbon: 6 12.01 = 72.06 Mass of hydrogen: 12 1.01 = 12.12 Mass of oxygen: 6 16 = 96 The percentage composition of carbon is: The percentage composition of hydrogen is: The percentage composition of oxygen is: Examples 3 What is the percentage composition of chromium in ? Molar mass of compound: 2 39 + 2 52 + 7 16 = 249 Molar mass of chromium: The percentage composition of Cr is: Examples 4 What is the percentage by mass of nitrogen in ammonium nitrate, NH4NO3 an important source of fertilizer? Molar mass of N: = 14.01 Molar mass of compound: 2 + 4 + 3 16 = The percentage composition of nitrogen is: The percentage composition of N is 35% Examples 5 Cinnamaldehyde, C9H8O, is responsible for the characteristic odour of cinnamon. Determine the percentage composition of cinnamaldehyde by calculating the mass percents of carbon, hydrogen, and oxygen. The molecular formula of cinnamaldehyde is C9H8O. Molar mass of C: = 12.01
16. 16. Page 16 Molar mass of H: = 1.01 Molar mass of O: = 16.00 First we calculate the molar mass of cinnamaldehyde: 12.01 + 8 1.01 + 16 = 132.17 Mass of carbon: = 9 12.01 = 108.09 Mass of hydrogen: = 8 1.01 = 8.08 Mass of oxygen: = 16 The percentage composition of carbon is: The percentage composition of hydrogen is: The percentage composition of oxygen is:
17. 17. Page 17 Determining empirical and molecular formulas The empirical formula is the simplest formula for a compound. A molecular formula is the same as or a multiple of the empirical formula, and is based on the actual number of atoms of each type in the compound. For example, if the empirical formula of a compound is C3H8 , its molecular formula may be C3H8 , C6H16 , etc. An empirical formula is often calculated from elemental composition data. The weight percentage of each of the elements present in the compound is given by this elemental composition. Using basic mathematics skills like ratio, percentage, linear equations and system of linear equations we can determine the empirical formula of an unknown compound from its atomic masses and percent composition. Example 1 Analysis of a compound gives 30.43 % N and 69.57% O. The mass for this compound is 92u. What is its molecular formula? Given Find Formula: %N = 30.43% %O = 69.57% Molar mass of N: = 14 Molar mass of O: = 16 First method We assume that the molecular weight is 100% and calculate the mass of nitrogen It means that 28 in the compound is for the atoms of nitrogen and the rest: 92 - 28 = 64 is for the oxygen. The numbers of atoms in the compound is: atoms of N atoms of O The formula is . Picture from: www.chem4kids.com
18. 18. Page 18 Second method In the second method we use the system of linear equations. First equation: Second equation: The formula molecular is . Empirical Formula Calculation Steps Step 1 If we have masses go onto step 2. If we have %. Assume the mass to be 100g, so the % becomes grams e.g. 40% of a compound is carbon. 40% of 100 g is 40 grams. Step 2 Determine the moles of each element. Step 3 Determine the mole ratio by dividing each elements number of moles by the smallest value from step 2. Picturefrom:www.commons.wikimedia.org
19. 19. Page 19 Step 4 Round your ratio to the nearest whole number as long as it is “close.” For example, 1.99987 can be rounded to 2, but 1.3333 cannot be rounded to 1. It is four-thirds, so we must multiply all ratios by 3 to rid ourselves of the fraction. If we have the empirical formula C1.5H3O1 we should convert all subscripts to whole numbers, multiply each subscript by 2. This gives us the empirical formula C3H6O2. Thus, a ratio that involves a decimal ending in .5 must be doubled. We should double, triple to get an integer if they are not all whole numbers. Example 2 A sulfide of iron was formed by combining 2.233 g of Fe with 1.926 g of S calculate the empirical formula. Molar mass of iron: = 55.85 Molar mass of sulfur: = 32.1 1 Mass of iron: = 2.233g Mass of sulfur: = 1.926g 2 Convert masses to amounts in moles Numbers of moles of iron: Numbers of moles of sulfur: 3 Divide these numbers of moles by the smallest number (0.03998 in this case) Fe ⇒ S ⇒ Preliminary formula is:  Now we should multiply to get a whole number. In order to turn 1.5 into a whole number, we need to multiply by 2 – therefore all results must be multiplied by 2. The simplest formula is: Example 3 Find the empirical formula for a compound containing 36.5% sodium, 25.4% sulfur and 38.1% oxygen. Molar mass of sodium Na: = 23 Molar mass of sulfur S: = 32.1 Molar mass of oxygen O: = 16 ·2
20. 20. Page 20 1 We assume that we have 100g of total material, and % becomes grams Mass of sodium: = 36.5g Mass of hydrogen: = 25.4g Mass of oxygen: = 38.1g 2 Convert masses to amounts in moles Numbers of moles of sodium: Numbers of moles of sulfur: Numbers of moles of oxygen: 3 Divide these numbers of moles by the smallest number (0.79 in this case) Na ⇒ S ⇒ O ⇒ The formula is  sodium sulfite. Example 4 The composition of ascorbic acid (vitamin C) is 40.92% carbon, 4.58% hydrogen, and 54.50% oxygen. What is the empirical formula for vitamin C? Molar mass of carbon C: = 12.01 Molar mass of hydrogen H: = 1.01 Molar mass of oxygen O: = 16 1 We are only given mass %, and no weight of the compound so we will assume 100g of the compound, and % becomes grams Mass carbon = 40.92g Mass hydrogen = 4.58g Mass oxygen = 54.5g Picturefrom:www.hlylhg.com
21. 21. Page 21 Picture from:www:rippedclub.net 2 Convert masses to amounts in moles numbers of moles of carbon: numbers of moles of hydrogen: numbers of moles of oxygen: 3 Divide these numbers of moles by the smallest number (3.4 in this case) C ⇒ H ⇒ 4.533.4=1.33 O ⇒ 3.43.4=1 Preliminary formula is:  Multiply to get a whole number. In order to turn 1.33 into a whole number, we need to multiply by 3 – therefore all results must be multiplied by 3 The simplest empirical formula of vitamin C is C3H4O3 Example 5 Muscle soreness from physical activity is caused by a buildup of lactic acid in muscle tissue. Analysis of lactic acid reveals it to be 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Its molar mass is 90.088 . What are the empirical and molecular formulas? Molar mass of carbon C: = 12.01 Molar mass of hydrogen H: = 1.01 Molar mass of oxygen O: = 16 1 We are only given mass %, and no weight of the compound so we will assume 100g of the compound, and % becomes grams Mass carbon = 40g Mass hydrogen = 6.7g Mass oxygen = 53.3g 2 Convert masses to amounts in moles numbers of moles of carbon: ·3
22. 22. Page 22 numbers of moles of hydrogen: numbers of moles of oxygen: 3 Divide these numbers of moles by the smallest number C ⇒ H ⇒ O ⇒ The empirical formula of lactic acid C is C1H2O1 4 Next we calculate the empirical formula weight: 12.01 + 21.01+ 16 = 30.03 5 Divide the molecule weight 90.08 by the empirical formula weight The molecular formula of lactic acid is C3H6O3. Example 6 A compound is found to contain 50.05 % sulfur and 49.95 % oxygen by weight. The molecular weight for this compound is 64.07 . What is its molecular formula? Molar mass of sulfur S: = 32.1 Molar mass of oxygen O: = 16 1 First we will find the empirical formula. We assume 100 g of the compound is present and change the percents to grams: Mass sulfur = 50.05g Mass oxygen = 49.95g 2 Then we convert the masses to moles: Numbers of moles of sulfur: Numbers of moles of oxygen: 3 Divide by the lowest, seeking the smallest whole-number ratio: S ⇒ O ⇒ 4 And now we can write the empirical formula: SO2 5 Next we calculate the empirical formula weight 32 + 216 = 64
23. 23. Page 23 6 Divide the molecule weight by the empirical formula weight 7 Use the scaling factor computed just above to determine the molecular formula: SO2 times 1 gives SO2 for the molecular formula. Example 7 An analysis of nicotine, an addictive compound found in tobacco leaves, shows that it is 74.0% C, 8.65% H, and 17.35% N. Its molar mass is 162 g/mol. What are its empirical and molecular formulas? Molar mass of carbon C: = 12.01 Molar mass of hydrogen H: = 1.01 Molar mass of nitrogen N: = 14 1 We assume 100g of the compound, and % becomes grams Mass carbon = 74g Mass hydrogen = 8.65g Mass nitrogen = 17.35g 2 Convert masses to amounts in moles Numbers of moles of carbon: Numbers of moles of hydrogen: Numbers of moles of nitrogen: 3 Divide by the lowest, seeking the smallest whole-number ratio: C ⇒ O ⇒ N ⇒ 4 And now we can write the empirical formula: 5 Next we calculate the empirical formula weight 5·12.01 + 71.01+ 14 = 81.12
24. 24. Page 24 Picture from:www.previewcf.turbosquid.com 6 Divide the molecule weight by the empirical formula weight The molecular formula of nicotine is .
25. 25. Page 25 Concentration by percent Medicated syrup is an example of the concentration. Brine and syrup Brine is a concentrated solution of sodium chloride in water. In Poland we use brine to prepare our special cucumbers - gherkins. Syrup is a concentrated solution of sugar in water. Example of syrup is fruit syrup Calculating concentration of a chemical solution requires basic math skills like knowing percentage, equations or system of equations. Solutions are homogeneous mixtures of solute and solvent.  Solvent - the most abundant substance in a solution. In a liquid solution, the solvent does the dissolving.  Solute - the other substance in a solution. In a liquid solution, the solute is dissolved. Concentration refers to the amount of solute that is dissolved in a solvent. The concentration of a solution in percent can be expressed in two ways: as the ratio of the volume of the solute to the volume of the solution or as the ratio of the mass of the solute to the mass of the solution. To calculate percent concentration we use a formula: The mass percent of a solution is a way of expressing its concentration. Mass percent is found by dividing the mass of the solute by the mass of the solution and multiplying by 100; e.g. a solution of NaOH that is 28% NaOH by mass contains 28 g of NaOH for each 100 g of solution. Picture from: www.styl.pl
26. 26. Page 26 Example 1: Calculate concentration of sugar in compote. For every kilogram of fruits you need syrup made from 0.4 liter of water and 0.4 kilogram of sugar. Given Calculate mwater = 0.4L= 400g Cp = ? mfruit = 0.4kg= 400g msyrup = 400g+ 400g= 800g Answer: Concentration of syrup is 50%. Example 2: We mingled 200L milk which contain 2% butterfat and 50L milk which contain 4% butterfat. Calculate final percent concentration in milk. Given: Calculate Cp = ? 1 2 Answer: We got 2.4% milk. Example 3 A bottle of the antiseptic hydrogen peroxide H2O2 is labeled 3%. How many mL H2O2 are in a 473 mL bottle of this solution? Given: Calculate Cp = 3% 1 Answer: There are about 14.2mL of H2O2 in a bottle of this solution. Picturefrom:www.net/human-medications-to-give-at-home
27. 27. Page 27 Example 4 How many grams of salt do you need to make 500 grams of a solution with a concentration of 5% salt? Given Calculate Cp = 5% msolute = x ? msolution =500g msolute = x = 25g Answer: We need 25g of salt. Example 5 How many grams of water must be evaporated from 10 grams of a 40% saline solution to produce a 50% saline solution? Given Calculate x the amount of evaporated water ( in grams) The amount of salt in the beginning and after the evaporation of water is the same, Answer: 2 grams of water must be evaporated from the 40% saline solution. Picturefrom:www.blog.farwestclimatecontrol.com In the beginning In the end 10g of solution 40%·10 the amount of salt (10-x) of solution 50%·(10-x) the amount of salt x g of water
28. 28. Page 28 Example 6 How many grams of salt must be added to 30kilos of a 10% salt solution to increase the salt concentration to 25%. How many kilos of salt were added? Given Calculate x the amount of added salt ( in kilos) The amount of water in the beginning and after adding salt is the same. Answer: 0.6 kilograms of salt must be added to the solution to increase the salt concentration. To solve word problems involving percent concentration amounts, knowledge of solving systems of equations and percents is necessary. Below there are some percent concentration problems involve solving systems of equations when mixing two liquids with differing percent concentration amounts. Example 7 A 16% salt solution is mixed with a 4% salt solution. How many milliliters of each solution are needed to obtain 600 milliliters of a 10% solution? The table below help us organise information. Amount of solution (in mL) Percent Total 16% salt solution x 0.16 0.16x 4% salt solution y 0.4 0.4y mixture x + y = 600 0.10 0.10·600 = 60 The liters of salt solution from the 16% solution, plus the liters of acid in the 30% solution, add up to the liters of acid in the 10% solution The first column is for the amount of each item we have. In the beginning In the end 30kg of solution 10%·30 the amount of salt 90%·30 the amount of water (30+x) of solution 25%·(30+x) the amount of salt 75%·(30+x) the amount of salt x g of salt
29. 29. Page 29 Answer: 300mL of these solutions are needed to obtain 600 milliliters of a 10% solution. Example 8 How much 10% sulfuric acid (H2SO4) must be mixed with how much 30% sulfuric acid to make 200 milliliters of 15% sulfuric acid? Let's organise the information in the table Volume Percent Amount of salt 10% sulfuric acid x 0.10 0.1x 30% sulfuric acid y 0.30 0.3y mixture x + y = 200 0.15 0.15·200 = 30 Now that the table is filled, we can use it to get two equations. The "volume" and "amount of acid" columns will let us get two equations. Since x + y = 600, then x = 600 – y. We can substitute for x in our second equation, and eliminate one of the variables Since x + y = 200, then x = 200 – y. The liters of sulfuric acid from the 10% solution, plus the liters of acid in the 30% solution, add up to the liters of acid in the 15% solution The first column is for the amount of each item we have.
30. 30. Page 30 Answer: 150mL 10% sulfuric acid must be mixed with 50mL 30% sulfuric acid to make 200 milliliters of 15% sulfuric acid. We can substitute for x in our second equation, and eliminate one of the variables
31. 31. Page 31 Links: www.en.wikipedia.org www.bbc.co.uk/schools/gcsebitesize/science/ www.towson.edu
32. 32. Page 32 G I M N A Z J U M I M . A N N Y W A Z Ó W N Y , G O L U B - D O B R Z Y Ń