Clinical Chemistry - 02 Laboratory Mathematics

o CLINICAL CHEMISTRY LABORATORY MATHEMATICS UNIT CONVERSION • • • • • 2 types of Units of Measure used in Clinical Chemistry: Systeme Internationale d’Unites (SI units) • Based on metric system • Devised to provide a global uniform method of describing physical quantities. • Preferred unit of measure in clinical laboratories and scientific literatures. • Includes 7 basic units: meter (length), kilogram (mass), mole (substance) • Uses Standard Prefixes to express gravity of basic unit: milli-, centi-, micro-, etc. • SI recommends that laboratory reports are expressed in terms of substance concentration (moles) or mass of a substance (g, mg, mmol) per volume of solution (moles/L, g/dL, mg/dL, mmol/L) Conventional Units (US Conventional System) o Widely used unit of measure in the United States. o Includes units such as: ▪ Distance: inch, feet, miles, yard ▪ Area: square foot ▪ Volume: cubic inch, tablespoon, pint, ▪ Mass: pound, ounce, 16 oz. = 1 lb. 7 Basic Units (SI Unit) BASE NAME SYMBOL Length Meter m Mass Kilogram kg Time Second s Electric Current Ampere A Thermodynamic temperature Kelvin K Amount of Substance Mole mol Luminous intensity Candela cd Standard Prefixes 2. • • 1. Move the decimal by the difference between the exponents represented by the prefix expressed in the multiplying factor. Sample Problem 2: Convert 2000 uL (microliter) to mL u = 10-6 -6 - (-3) m = 10-3 -6 + 3 = -3 → L 2,000 x 10-3 mL = 2.0 mL CONVERSION IN CLINICAL CHEMISTRY ANALYTES • • • SI UNIT CONVERSION Conversion (smaller to larger): move decimal to the left. ▪ Mill → deka 10-3 → 101 (-3) -1 = 10-4 o Conversion (larger to smaller): move decimal to the right. o Negative difference to left o Positive difference to right Write the correct symbol after the numerical value. Sample Problem 1: Convert 1 m to um (micrometer) m = 100 0 - (-6) u = 10-6 0+6 = 106 → R 1.0 x 106 um = 1,000,000 um ALBUMIN AMMONIA BICARBONATE BILIRUBIN BUN CALCIUM CHLORIDE CHOLESTERTOL CREATININE GLUCOSE IRON LITHIUM MAGNESIUM PHOSPHOLIPID PHOSPHORUS POTASSIUM SODIUM THYROXINE TOTAL PROTEIN TRIGLYCERIDE URIC ACID pCO2 pO2 CONVENTIONAL UNIT g/dL μg/dL mEq/L mg/dL mg/dL mg/dL mEq/L mg/dL mg/dL mg/dL mg/dL mEq/L mEq/L g/dL mg/dL mEq/L mEq/L μg/dL g/dL mg/dL mg/dL mm/Hg mm/Hg SI UNIT g/L μmol/L mmol/L μmol/L mmol/L mmol/L mmol/L mmol/L μmol/L mmol/L μmol/L μmol/L mmol/L g/L mmol/L mmol/L mmol/L nmol/L g/L mmol/L mmol/L pKa pKa CONVERSION FACTOR 10 0.587 1.0 17.1 0.357 0.25 1.0 0.026 88.4 0.0555 0.179 1.0 0.5 0.01 0.323 1.0 1.0 12.9 10 0.0113 0.0595 0.133 0.133 Conventional → SI = multiply (x) SI → Conventional = divide (/) Sample Problem 1: What is 95 mg/dL of glucose in mmol/L? glu = 0.0555 mg/dL → mmol/L = x 95 x 0.0555 = 5.27 mmol/L • Sample Problem 2: Convert 0.95 mg/dL creatinine to umol/L. Creatinine = 88.4 mg/dL → umol/L = x 0.95 x 88.4 = 84.0 umol/L • Sample Problem 3: What is 6.01 mmol/L of glucose in mg/dL? glu = 0.0555 mmol/L → mg/dL = / 6.01 / 0.0555 = 108.29 mg/dL PERCENT SOLUTION • • • Percent – “parts per 100” Percent Solution – amount of solute per 100 total units of solution Weight/Volume (w/v) % Solution o Amount of solute in grams per 100 mL of solvent o Most commonly used in clinical lab • Volume/Volume (v/v) % Solution o Amount of Solute in mL in 100 mL of solvent • Weight/Weight (w/w) % Solution o Amount of solute in grams per 100 grams of solution MOLARITY (M) • • • • SAMPLE PROBLEMS 1. w/v%: How many grams of HCl is needed to make a 250 mL of 5% HCl solution? Number of moles of solute per liter of solution. One mole of substance is equivalent to its gram molecular weight (gmw). GMW - sum of the atomic weights of the component element of solute FORMULA: M= moles of solute . Volume of Solution (L) o To get mole of solute: moles = grams of solute . Gram Molecular Weight (GMW) o Example: ▪ What is the GMW of CuSO4? Cu – 63.5 S – 32.1 O – 16.0 * 4 = 64 GMW = 159.6 o To convert w/v % solution to Molarity: M = (M/V %)(10) GMW o To convert Molarity to w/v % solution: W/V % = (M)(GMW) 10 SAMPLE PROBLEM 2. 3. 1. 250 grams of NaCl dissolved in 250 mL of distilled water will yield a Molarity of? (Na: 23.0; Cl: 35.5) 2. How many grams of Na2CO3 is needed to yield 400mL of 0.12M Na2CO3 solution? (Na: 23.0; C: 12.0; O: 16.0) 3. 4. Compute for the molarity of 0.85% w/v NaCl solution What is 2.5 M H2SO4 solution in %w/v? 3 4. v/v%: How many mL of NaOH is needed to make a 50 mL of 2% NaOH solution? w/w%: How many grams of CuSO4 · 5H2O is needed to make a 1000 grams of 7% aqueous solution Fehlings A reagent? MOLALITY (m) • • • • • Number of moles of solute per kilogram of solvent. SI Unit = mol/kg. Depend only with the mass of the solvent rather than solution FORMULA: m = moles of solute . Kilogram of solvent o To get mole of solute: o moles = grams of solute . Gram Molecular Weight (GMW) Sample Problem: Calculate the molality of 15.0 grams of KBr dissolved in 500 mL of distilled water? (K: 39.1; Br: 79.9) SAMPLE PROBLEM • Calculate the normality of a 500 mL solution that contains 7 g of H2SO4? (H: 1.0; S: 32.1; O: 16.0) • What is the molarity of a 2.5 N solution of HCl? • What is the Normality of a 0.5 M solution of H2SO4? • How many grams of Al(OH)3 is needed to yield 250 mL of 0.08 N Al(OH)3 solution? (Al: 27.0; O: 16.0; H: 1.0) NORMALITY (N) • • • • Number of equivalent weight of solute per liter of solution Often been used in acid-base calculations. FORMULA: N= grams of solute . (Equivalent weight)(volume of solution in L) o To get equivalent weight (EW) of solute: EW = GMW Valence o Valence ▪ Acid = # of H+ ions that can be donated HCl, HNO3 - 1 valence H2SO4 - 2 valence H3PO4 - 3 valence ▪ Base = # of replaceable OH- ions NH4OH, KOH, NaOH - 1 valance Na(OH)2, Ba(OH)2 - 2 valence Al(OH)3 - 3 valence ▪ Salts - # of replaceable anion NaCl - 1 valence Na2CO3 - 2 valence o To convert w/v % solution to Normality: N = (W/V%)(10) EW o To convert Normality to w/v % solution: o W/V % = (N)(EW) 10 Relationship between Molarity and Normality: o Normality = (Molarity)(Valence) o Molarity = Normality Valence DILUTION • • • • Ratio of concentrated or stock material to the total final volume of the solution. It consist of the volume or weight of the concentrate plus the volume of the diluent. Dilution Factor (DF) – ratio of concentrated or stock solution to the total solution volume. o Inversely proportional to the concentration of the solution. ▪ ↓DF = less concentration ▪ ↑DF = more concentration o FORMULA: DF = final concentration initial concentration Involves two entities: o Solute – substance being diluted o Diluent – medium making up the rest of the solution. • • Dilution = Amount of solute Volume of solution or Amount of solute : total volume of solution Solution = solute + diluent Types of Dilution: o Simple Dilution o Compound Dilution o Serial Dilution SERIAL DILUTION • • • SIMPLE DILUTION • Series of dilution in which the original dilution is further diluted using same dilution factor in each step of dilution. Often used to obtain titer (antibody strength) – the reciprocal of the highest dilution of serum that gives positive reaction with antigen. Sample: A set-up of six test tubes with 0.2 mL of diluent each. If 0.2 mL of serum is added to the first tube, this becomes a 1:2 dilution. 2mL of a 1:20 dilution is needed to run a specific serological test. How much serum and how much diluent are needed to make this dilution? o Dilution: 1:20 or 1/20 o Volume of Solution: 2 mL o Amount of solute (serum): ? (x) = 0.1 mL o Amount of diluent: ? (y) = 1.9 mL 1/16 = highest dilution that gives + reaction • A 1:5 dilution of patient serum is necessary to run a serological test. There is 0.1 mL of serum that can be used. What amount of diluent is necessary to make this dilution using all of the serum? o Dilution: 1:5 or 1/5 o Volume of Solution: ? (x) = 0.5 mL o Amount of solute (serum): 0.1 mL o Amount of diluent: ? (y) = 0.4 mL pH & pOH pH • • • • COMPOUND DILUTION • • • Process of yielding a large dilution by doing a series of dilution using the preceding diluted solution to attain the desired dilution. Advantages: More accurate and less costly. FDpreceding tube x ID of the next tube Sample: How much serum would it take to have 1:500 dilution? negative log of the hydrogen ion concentration Scales from “0-14” – the value expresses hydrogen ion concentration. o <7.0 = acidic o 7.0 = neutral o >7.0 = basic Mathematically expressed as: o pH = − log H+ o To calculate the pH, you need to know the concentration of hydrogen ion in mol/L (Molarity). Sample Problem: What is the pH of 0.25 mol/L of HCl? pH = -log (H+) pH = -log (0.25) pH = -(-0.60) pH = 0.60 pOH • • • • negative log of the hydroxide ion concentration Used to express alkalinity of a solution. Mathematically expressed as: o pOH = − log OH− Sample Problem: What is the pOH of 0.62 mol/L of NaOH? pOH = -log (OH-) pOH = -log (0.62) pOH = -(-0.21) pOH = 0.21 pH and pOH 22mL diluent • • pH + pOH = 14.0 Sample Problem: What is the pOH of 0.25 mol/L of HCl? pH + pOH = 14.0 pH = -log (H+) pOH = 14 - pH pH = -log (0.25) pOH = 14 - 0.6 pH = -(-0.61) = 0.6 pOH = 13.4