Stoichiometry

Methanol, CH3OH, which is used as fuel in racing cars and fuel in cells, can be made by the reaction of carbon monoxide and hydrogen: CO(g) + 2H2(g) -> CH3OH(l) Suppose 356 g CO and 65.0 g H2 are mixed and allowed to react. (a) What mass of methanol can be produced? (b) What mass of the excess reactant remains after the limiting reactant has been consumed?

(a) Determine how many moles are in each sample: 365 g CO x (1 mol CO/28 g CO) = 12.7 mol CO (lim reactant) 65 g H2 x (1 mol H2/2 g H2) = 32.5 g H2 (excess) Use limiting reactant to determine g CH3OH: 12.7 mol CO x (1 mol CH3OH/1 mol CO) x (32 g CH3OH/1 mol CH3OH) = 406 g CH3OH (b) Mass is conserved across reactants and products. 356 g CO + 65.0 g H2 = 421 g in reactants However, only 406 g are produced 421 g - 406 g = 15 g not used -> must be H2 (See iPad for another explanation)

What is 2 moles of silicon equal to?

1 mole Si = 28 g 2 moles Si = 56 g

If 27.9 g of Fe great with 24.1 g of S to produce FeS what would be the limiting reagent? How many grams of excess reagent would be present in the vessel at the end of the reaction? The balanced equation is Fe + S -> FeS A. Fe is the limiting reagent with 8 g S as excess B. Fe is the limiting reagent with 12 g S as excess C. S is the limiting reagent with 8 g Fe as excess D. S is the limiting reagent with 12 g Fe as excess

1. Determine the number of moles for each reactant. 27.9 Fe x (1 mol Fe/55.8 g Fe) ~ 28 x 1/56 = 0.5 Fe 24.1 g S x (1 mol S/32.1 g S) ~ 24 x 1/32 = 0.75 mol S 2. Because 1 mole of Fe is needed to react with 1 mole of S and there are 0.5 moles Fe for the given 0.75 moles S, the limiting reagent is Fe. Thus, 0.5 moles of Fe will react with 0.5 moles of S, leaving an excess of 0.25 moles of S in the vessel. The mass of the excess reagent will be: 0.25 mol x (32.1 g S/1 mol S) = 8 g S

What are the empirical and molecular formulas of a carbohydrate that contains 40.9% carbon, 4.58% hydrogen, and 54.52% oxygen and has a molar mass of 264 g/mol? A. Empirical: C3H4O3, Molecular: C6H8O6 B. Empirical: C3H4O3, Molecular C9H12O9 C. Empirical: C7H8O7, Molecular: C14H16O14 D. Empirical: C7H8O7, Molecular: C7H8O7

1. Determine the number of moles of each element in the compound by assuming a 100 g sample; this converts the percentage of each element by present directly into grams of that element. Then converts grams to moles: moles C = 40.9 g/(12 g/mol) ~ 3.4 mol moles H = 4.58 g/(1 g/mol) ~ 4.6 mol moles O ~ 54.52 g/(16 g/mol) ~ 3.4 moles 2. Find the simplest whole number ratio of the elements by dividing the number of moles for each compound by the smallest number out of all obtained in the previous step. C: 3.4/3.4 = 1.00 H: 4.6/3.4 ~ 1.33 O: 3.4/3.4 ~ 1.00 3. The empirical formula is obtained by converting the numbers obtained into whole numbers by multiplying them in an integer value. Empirical formula = C1H1.33O1 x 3 = C3H4O3 4. To determine molecular formula, divide the molar mass (264 g/mol, given in the question stem) by the empirical formula weight. The resulting value gives the number of empirical formula units in the molecular formula The formula weight of the empirical formula C3H4O3 is: (3x12.0 g/mol C) + ( 4 x1.0 g/mol H) + (3x16.0 g/mol O) = 88 g/mol total ratio of molecular formula/empirical formula - 264/88 = 3 5. Find the molecular formula by multiplying by this ratio C3H4O3 X 3 = C9H12O9

How many molecules of H2O are in 20 g of water?

20 g H2O x (1 mol/18 g H2O) = 1.11 moles H2O 1.11 moles H2O x (6.02 x 10^23 molecules/1 mole) = 6.68 x 10^23 molecules of H2O

Identify the limiting reagent, and find the mass of the excess reagent left over once the reaction has run to completion. (46 g Na and 32 g O2 are provided) 4Na(s) + O2(g) -> 2Na2O(s) A. Sodium, 11 g O2 B. Sodium, 16 g O2 C. Oxygen, 11 g Na D. Oxygen, 16 g Na

46 g Na x (1 mol Na/23 g Na) = 2 mol Na 32 g O2 x (1 mol O2/32 g O2) = 1 mol O2 Because 4 sodium atoms are needed for every oxygen molecule, sodium will run out first. To determine the amount of Na2O formed: 2 mol Na x (2 mol Na2O/4 mol Na) = 1 mol Na2O 2 mol Na x (1 mol O2/4 mol Na) = 0.5 mol O2 will be used, so 1.0-0.5 mol O2 = 0.5 mol O2 will remain. In grams this is: 0.5 mol O2 x (32 g O2/1 mol O2) = 16 g excess O2 Answer choice B

How many atoms of H2O are in 20 g of water?

6.68 x 10^23 molecules of H2O x 3 = 20.04 x 10^23 atoms Because there are 3 atoms (2 hydrogens and 1 oxygens) in a water molecule

In the reaction shown, if 39.05 g of Na2S are reacted with 85.5 g of AgNO3, how much of the excess reagent will be left over once the reaction has gone to completion? Na2S + 2 AgNO3 -> Ag2S + 2 NaNO3 A. 19.5 g Na2S B. 26.0 g Na2S C. 41.4 g AgNO3 D. 74.3 g AgNO3

A

In which of the following compounds is the percent composition of carbon by mass closest to 62%? A. Acetone B. Ethanol C. Propane D. Methanol

A

Using a given mass of KClO3, how would one calculate the mass of oxygen produced in the following reaction, assuming it goes to completion? 2 KClO3 → 2 KCl + 3 O2 A. (grams KClO3 consumed)(3 moles O2)(molar mass O2) / (molar mass KClO3)(2 moles KClO3) B. (grams KClO3 consumed)(molar mass O2) / (molar mass KClO3)(2 moles KClO3) C. (molar mass KClO3)(2 moles KClO3) / (grams KClO3 consumed)(molar mass O2) D. (grams KClO3 consumed)(3 moles O2) / (molar mass KClO3)(2 moles KClO3)(molar mass O2)

A

Which of the following compounds has a formula weight between 74 and 75 grams per mole? A. KCl B. C4H10O C. MgCl2 D. BF3

A

Which of the following is the gram equivalent weight of H2SO4 with respect to protons? A. 49.1 g B. 98.1 g C. 147.1 g D. 196.2 g

A

What is the molecular weight of SOCl2?

Add together the atomic weights of each of the atoms 1 S: 1 x 32.1 amu = 32.1 amu 1 O: 1 x 16.0 amu = 16.0 amu 2 Cl: 2 x 35.5 amu = 71.0 amu Total molecular weight = 119.1 amu per molecule

Calculate the molar mass and number of moles in 100 g of SrCl2. A. 158.6 g/mol, 0.625 moles B. 512.9 g/mol, 0.942 moles C. 893.4 g/mol, 0.182 moles D. 216.7 g/mol, 0.476 moles

Answer choice A SrCl2: 87.6 + (2 x 35.5) = 87.6 + 71 = 158.6 g/mol 100 g/158.6 g/mol ~ 100/160 = 10/16 = 5/8 = 0.625 moles Note the answer is a fraction based on eighths. As these are commonly used on the MCAT, knowing the values for 1/8 through 7/8 can be useful for many problems

Calculate the molar mass and number of moles in 100 g of NaBr. A. 103 g/mol, 0.97 moles B. 212 g/mol, 0.62 moles C. 344 g/mol, 0.77 moles D. 209 g/mol, 0.34 moles

Answer choice A. NaBr: 23 + 79.9 ~ 23 + 80 = 103 g/mol 100 g/103 g/mol = 0.97 moles Note that the denominator is greater than the numerator by approximately 3%, and the actual value is less than 1 by approximately 3%. This approximation can be used to quickly estimate answers that are close to one.

Calculate the molar mass and number of moles in 100 g of C6H12O6. A. 280 g/mol, 0.468 moles B. 180 g/mol, 0.555 moles C. 533 g/mol, 0.461 moles D. 570 g/mol, 0.253 moles

Answer choice B C6H12O6: (6 x12) + (12 x 1) + (6 x 16) = 180 g/mol 100 g/180 g/mol = 10/18 = 5/9 = 0.555 moles Note the final fraction is in ninths. Dividing by nine follows a standard pattern that is useful to know for the MCAT: 1/9 = 0.111, 2/9 = 0.222, 3/9 = 0.333, etc.

Assume you have a sample of 100 g of Hg and Cl where 73% is Hg and 27% is Cl. Estimate the empirical formula of Hg and Cl based off their atomic weights.

Approximately 0.364 moles Hg and 0.762 moles Cl. 0.762 is approximately 2 times more than 0.364 so the empirical formula is approximately HgCl2. (See iPad for worked explanation)

In the following reaction: Au2S3(s) + H2 (g) -> Au(s) + H2S(g) If 2 moles of Au2S3(s) is reacted with 5 moles of hydrogen gas, what is the limiting reagent? A. Au2S3(s) B. H2(g) C. Au(s) D. H2S(g)

B

In the process of photosynthesis, carbon dioxide and water, combine with energy to form glucose and oxygen, according to the following equation: CO2 +H2O -> C6H12O6 + O2 What is the theoretical yield of glucose if 30 grams of water are reacted with excess carbon dioxide and energy, according to the equation above? A. 30.0 g B. 50.0 g C 300.1 g D. 1801 g

B

If you are given 34 g NH3 and 32 g O2, how many grams of NO are produced? NH3 + O2 -> NO + H2O

Balance equation: 4NH3 + 5O2 -> 4NO + 6H2O Determine how many moles are in each sample: NH3 has a molar mass of 17 g = 2 mol NH3 in 34 g O2 has a molar mass of 32 g = 1 mol O2 in 32 g (limiting reactant) 32 g O2 x (1 mol O2/32 g O2) x (4 mol NO/5 mol O2) x (30 g NO/1 mol NO) = 24 g NO

If 46 g Na and 32 g O2 are provided, find the maximum number of moles of sodium oxide produced. Na(s) + O2(g) -> Na2O(s)

Balance the equation: 4Na(s) + O2(g) -> 2Na2O(s) 46 g Na x (1 mol Na/23 g Na) = 2 mol Na 32 g O2 x. (1 mol O2/32 gO2) = 1 mol O2 Because 4 sodium atoms are needed for every oxygen molecule, sodium will run out first. To determine the amount of Na2O formed: 2 mol Na x (2 mol Na2O/4 mol Na) = 1 mol Na2O

If you have 5 grams of sucrose (342 g), how many grams of oxygen are needed for the reaction? C12H22O11 + O2 -> CO2 + H2O

Balance the equation: C12H22O11 + 12O2 -> 12CO2 + 11H2O 5 g sucrose x (1 mol sucrose/342 g sucrose) x (12 mol O2/1 mol sucrose) x (348 g O2/1 mol O2) = 5.6

If there are 1.45 g of P4(s), how many grams of Cl2 are required? How many PCl3 produced? P4(s) + Cl2(aq) -> PCl3(l)

Balance the equation: P4(s) + 6Cl2(aq) -> 4PCl3(l) 31 g P x 4 = 124 g P4 1.45 g P4 x (1 mol P4/124 g P4) x (6 mol Cl2/1 mol P4) x (70.906 g Cl2/1 mol Cl2) = 4.96 g Cl2 Mass is conserved on the reactants and products: 1.45 g P4 + 4.96 g Cl2 = 6.41 g PCl3

How do the number of molecules in 18 g of H2O compare to the number of formula units in 58.5 g of NaCl?

Both values equal one mole of the given substance. The number of entities in a mole is always the same (Avogadro's number, 6.022 x 10^23 mol^-1)

What is the molecular formula of a compound with an empirical formula of B2H5 and a molar mass of 53.2 g/mol? A. B2H5 B. B3H7 C. B4B10 D. B6H15

C

What do the coefficients of a reaction represent? (# of 3 things) What don't they represent?

Coefficients represent # moles, molecules, and atoms but NOT the mass of any kind

Which of the following best describes ionic compounds? A. Ionic compounds are formed from molecules containing two or more atoms. B. Ionic compounds are formed of charge particle and are measured by molecular weight C. Ionic compounds are formed of charged particles that share electrons equally. D. Ionic compounds are three-dimensional arrays of charged particles

D

What type of formula do ionic compounds have, such as NaCl and CaCO3?

Empirical formulas

True or false: You can determine molecular formula from empirical formula.

False; you can only determine empirical from molecular Ex. C6H6 can be shown as the empirical formula CH but you are unable to determine that C6H6 came from CH because many other molecules have the same 1:1 ratio

What is the gram equivalent weight (GEW) of sulfuric acid?

Find the molar mass of H2SO4: (2 x 1.0 g/mol H) + (1 x 32.1 g/mol S) + (4 x 16.0 g/mol O) = 98.1 g/mol H2SO4 Next, identify the equivalents: protons (H+), because these are transferred in acid-base reactions. The number of protons in sulfuric acid (n) is 2. Now, calculate the gram equivalent weight: gram equivalent weight = molar mass/n GEW = (98.1 g/molH2SO4)/(2 mol H+/mol H2SO4) = 49.05 g/mol H+

How many moles are in 9.53 g MgCl2?

Find the molar mass of MgCl2: (1 x 24.3 g/mol) + (2 x 35.5 g/mol) = 95.3 g/mol Solve for number of moles: 9.53 g x (1 mole/95.3 g) = 0.10 mol MgCl2

If you have 85 g of Fe2O3, how many grams of Aluminum are needed? Fe2O3 + Al -> Al2O3 + Fe

First balance the equation: Fe2O3 + 2Al -> Al2O3 + 2Fe For every 1 mole of Fe2O3, there are 2 Aluminums Determine molar mass of Fe2O3: 2x56 + 3x16 = 160 g 85 g x (1 mol/160 g) = 0.53 mol Fe2O3 Multiply 0.53 moles x 2 because 1 mol Fe2O3: 2 moles Aluminum = 1.06 moles Al 1.06 mol x (27 g/1 mol) = 28.62 g Al

What is the normality of a 2 M Mg(OH)2 solution?

First identify the number of equivalents (n). There are two hydroxide ions (OH-) for each molecule of Mg(OH)2, which is the equivalent of interest because magnesium hydroxide is a base. Then, calculate the normality. Normality - molarity x n = 2 M x 2 equiv OH-/mol M(OH)2 = 4 N Mg(OH)2

What conclusions can you make about moles of each reactant making moles of product in the equation? 2H2 + O2 -> 2H2O

For 1 mole of H2 gas consumed, 1 mol water can be produced For every 1 mol O2 consumed, 2 mol water can be produced Hydrogen gas is being consumed at a rate twice that of oxygen gas

Determine the normality of 95 g PO4 3- in 100 mL solution. (Note: The species of interest is H+) A. 30 N B. 79 N C. 33 N D. 91 N

For PO4 3- the grams must first be converted to moles then to normality. PO4 3- has a molecular mass of 95, giving: 95 g/100 mL = 950 g/1 L = 10 mol/1 L = 10 M PO4 3- 10 M PO4 3- x (3 equiv H+/mol PO4 3-) = 30 N PO4 3- Answer choice A

What does it mean for a reaction to "run to completion?" How come most reactions do not run to completion?

It means that one of the reactants are used up completely. Many reactions achieve equilibrium before any of the reactants are used up completely. X + Y <--> XY

What does stoichiometry refer to?

Measuring chemicals that go into and out of any given reaction. It allows us to count up atoms and molecules by weighing them.

How many grams of calcium chloride are needed to prepare 71.7 g of silver chloride according to the following equation? CaCl2(aq) + 2AgNO3(aq) -> Ca(NO3)2(aq) + 2AgCl(s) A. 27.8 B. 39.0 C. 89.2 D. 45.7

Molar mass of CaCl2 = 111.1 g Molar mass of AgCl = 143.4 g 71.7 g AgCl x (1 mol AgCl/123.4 g AgCl)(1 mol CaCl2/2 mol AgCl)(111.1 g CaCl2/1 mol CaCl2) Estimate: (72/144)x(1/2)x(110/1) = (1/2) x (1/2) x (110/4) = 27.5 CaCl2 (actual value 27.775) which is answer A

Find the percent composition by mass of sodium, carbon, and oxygen in a sodium carbonate molecule (Na2CO3) A. 31% Na, 6% C, 63% O B. 76% Na, 5% C, 19% O C. 46% Na, 12% C, 46% O D. 40% Na, 18% C, 42% O

Molar mass of sodium carbonate is given by (2 x 23) + (1 x 12) + (3 x 16) = 106 g/mol. Sodium: (2 x 23 g/mol)/106 g/mol x 100% ~ 46/100 = 46% Carbon: (1 x 12 g/mol)/106 g/mol x 100% ~ 12/100 = 12% Oxygen: (3 x 16 g/mol)/106 g/mol x 100% ~ 48/100 = 48% Answer choice C

Compare the units of molar mass and molecular weight

Molar mass refers to the mass of one moles of a compound which is expressed in g/mol Molecular weight is measured in amu/molecule

What is the empirical and molecular formula of water?

Molecular and empirical are both: H2O

What is the difference between molecular and empirical formula of a molecule?

Molecular formula: actual number of elemental atoms in a molecule; a multiple of the empirical formula Empirical formula: observed simplest whole number ratio of atoms in a molecule through experimentation

Compare and contrast molecular weight and formula weight.

Molecular weight: the sum of the atomic weights of all the atoms in a molecule Formula weight: adding up the atomic weights of the constituent ions according to its empirical formula Both are measured in amu per molecule

What is the molecular and empirical formula of benzene?

Molecular: C6H6 Empirical: CH

What is the empirical and molecular formula of octasulfur?

Molecular: S8 Empirical: S

What is the purpose of moles in regards to conversion?

Moles are a unit that convert between atomic mass units, grams, and atoms (6.02 x 10^23) 1 gram = 1 mole of amu's 1 gram = 6.02 x 10^23 amu

Can one mole of HCl completely neutralize one mole of Ca(OH)2? Why or why not?

No, one mole of HCl will donate one equivalent of acid but Ca(OH)2 will donate 2 equivalents of base

Are ionic compounds a molecule? Why or why not?

No, they do not form true molecules because of the way in which the opposite charged ions arrange themselves in the solid state. Molecules are held together by covalent bonds Ex. Solid NaCl is a coordinated lattice where Na+ ions are surrounded by Cl- atoms and vice versa. This makes it hard to clearly define a sodium chloride molecule

Determine the normality of 0.25 M H3PO4. (Note: The species of interest is H+) A. 0.11 N B. 0.75 N C. 0.86 N D. 0.68 N

Normality is calculated as N = M x equivalents/mole For H3PO4: 0.25 M H3PO4 x (3 equiv H+/mol H3PO4) = 0.75 N H3PO4 Answer choice B

What is the percent composition of chromium in K2Cr2O7? A. 40% B. 20% C. 38% D. 35.4%

Percent composition (of an element by mass): percent of a specific compound that is made up of a given element. Percent composition - mass of element in formula/molar mass x 100% Molar mass of K2Cr2O7: (2x39.1 g/mol) + (2x52.0 g/mol) + (7x16.0 g/mol) ~ (2x40) + (2x50) + (7x16) = 292 g/mol(actual value = 294.2 g/mol) Calculate the percent composition of Cr: percent composition = (2 x52.0 g/mol)/(294.2 g/mol) ~ 2x50/300 x 100% = 100/300 x 100% = 33% (actual value 35.4%) Answer choice D

What is the difference of between a physical and chemical reaction? Give examples of each.

Physical reaction: When a compound undergoes a reaction without changing its molecular formula Ex. melting, freezing, evaporating, condensation Chemical reaction: When a compound reacts to produce a different molecular formula Ex. redox, combination, addition, combustion of hydrocarbon

Be(OH)2 is produced when water reacts with BeO. Starting with 2.5 kg BeO in excess water, and producing 1.1 kg Be(OH)2, what is the percent yield of this reaction? A. 25.6% B. 26.4% C. 81.9% D. 53.0%

Reaction: BeO + H2O -> Be(OH)2 Theoretical yield: 2500 g BeO x (1 mol BeO/25 g BeO)(1 mol Be(OH)2/1 mol BeO)(43 g Be(OH)2/1 mol Be(OH)2) = 4300 g Be(OH)2 Percent yield = actual yield/theoretical yield x 100% = 1100 g/4300 g x 100% ~ 11/44 x100% = 25% yield (actual: 25.6%)

How many equivalents of electrons does Sodium and Magnesium produce, separately?

Sodium will donate one mole of electrons (one equivalent) Magnesium will donate two moles of electrons (two equivalents)

Experimental data from the combustion of an unknown compound indicates that it is 28.5% iron, 24.0% sulfur, and 49.7% oxygen by mass. What is its empirical formula? A. Fe2S3O12 B. Fe2S5O12 C. FeS2O8 D. Fe4S5O20

Start by assuming a 100 g sample, which represents 28.5 g Fe, 24.0 g S, and 49.7 g O. Next divide each number of grams by the atomic weight to determine the number of moles: Fe: 28.5 g/55.8 g/mol ~ 28/56 = 0.5 moles S: 24 g/32.1 g/mol ~ 24/32 = 3/4 = 0.75 moles O: 49.7 g/16 g/mol ~ 48/16 = 3 moles Next, find the multiplier that gives all three compounds integer values of moles. Sulfur: 0.74 x 4 = 3 moles Fe: 0.5 x 4 = 2 moles O: 3 x 4 = 12 moles Answer choice A

What does the law of constant composition say about a sample of water (regards to atoms and mass)?

The Law of constant compositions states that any pure sample of a given compound will contain the same elements in an identical mass ration. Every sample of water will contain 2 hydrogen atoms for every 1 oxygen atom. For every 1 g H will have 8 g O.

What is the percent yield for a reaction in which 28 g of Cu is produced by reacting 32.7 g of Zn in excess CSO4 solution? A. 23.7% B. 87.5% C. 49.8% D. 43.9%

The balanced equation is as follows: Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq) Calculate the theoretical yield for Cu 32.7 g Zn x (1 mol Zn/65.4 g Zn)(1 mol Cu/1 mol Zn)(63.5 g Cu/1 mol Cu) ~ 33 x (1/66) x (1/1) x (64/1) = 1/2 x 64/1 = 32 (actual value = 31.8 g Cu) This 31.8 represents the theoretical yield. Finally determine the percent yield. 28 g/31.8 g x 100% ~ 28/32 x 100% = 7/8 x 100% = 87.5%

In a 1 N HCl solution, what is the molarity of HCl? Molarity of H2CO3 in 1 N H2CO3?

The following equation can be used to determine molarity: molarity = normality/n - n = number of protons, hydroxide ions, electrons, or ions produced or consumed by the solute 1 M of HCl because HCl is a monoprotic acid 0.5 M H2CO3 because H2CO3 is a diprotic acid - 2 equivalents of base are required to neutralize both protons of the acid

What information does the formula C6H12O6 (glucose) tell us? What does it not tell us?

The particular compound consists of 6 atoms of carbon, 12 atoms of hydrogen, and 6 atoms of oxygen. There is no indication of how the different atoms are arranged or how many bonds exist between each of the atoms. Structural formulas show various bonds between the constituent atoms of a compound.

What is molecular mass?

The sum of the masses of all the atoms in a molecule

What is the difference between theoretical yield and actual yield? How do they both relate to percent yield?

Theoretical yield: the maximum amount of product that can be generated as predicted from the balanced equation, assuming that all of the limiting reactant is consumed, no side reactions have occurred, and the entire product has been collected. It is rarely ever attain ed through the actual chemical reaction Actual yield: the amount of product one actually obtains during the reaction Percent yield = actual yield/theoretical yield x 100%

CH4 + 2O2 --> CO2 + 2H2O If 2 moles of methane react with 6 moles of oxygen, how many moles of each product are produced? Are there any limiting reactants?

When this reaction runs to completion, 2 moles of methane and 4 moles of oxygen are reacted to form 2 moles of carbon dioxide and 4 moles of water. However, there are still 2 moles of oxygen that are not reacted. Therefore, methane is the limiting reactant.

What similar equation can you use the following equation when determining equivalents of a compound? gram equivalent weight = molar mass/n

equivalents = mass of compound (g)/gram equivalent weight (g)

How many grams of H2CO3 (62 g/mol) would you need to produce one equivalent of hydrogen ions?

gram equivalent weight = molar mass/n - n = number particles of interest produced or consumed per molecule of the compound in the reaction gram equivalent weight = 62 g/mol/2 = 31 grams (H2CO3 can donate 2 hydrogen ions)

Which is the empirical and molecular formula for glucose? C6H12O6 CH2O

molecular: C6H12O6 empirical: CH2O

What is 1 mole of Aluminum equal to?

~27 grams or 26.98 g