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TOPIC 9.
CHEMICAL CALCULATIONS III - stoichiometry.
Stoichiometric calculations. By combining a knowledge of balancing equations with the concept of the mole, it is possible to easily calculate the masses of all reactants required or products formed in a n y rea c tio n. T h ese c a lc u la tio n s a r e c a lle d S T O I C H IO M E T R IC CALCULATIONS. How this is done is again best illustrated by some examples.
Example 1. Heating potassium chlorate, KClO , produces oxygen and potassium 3 chloride. What mass of oxygen is obtained by quantitatively decomposing 3.00 g of potassium chlorate?
Start by writing a balanced formula equation for the reaction.
2KClO 3 v 2KCl + 3O 2
Then write down the relative numbers of molecules (or formula units for ionic
compounds) involved as shown in the equation (the stoichiometric coefficients ).
2 molecules v 2 “molecules” + 3 molecules
If each of these numbers of molecules were multiplied by the Avogadro number,
written as N (^) A for short, we would have
2 N (^) A molecules v 2 N (^) A “molecules” + 3 N (^) A molecules
As the Avogadro number of molecules of any substance is 1 mole of the substance, then the relative numbers of moles of all species in the equation would be
2 moles v 2 moles + 3 moles
It can be seen then that the coefficients in the balanced chemical equation can be interpreted as relative numbers of moles of species as well as the relative numbers of molecules. Usually the problem does not require calculation of the masses of all of the reactants and/or products in the reaction. More often only two species need to be involved in the calculation and they are referred to as the known (the reactant/product
for which the masses are given, KClO 3 in this example) and the unknown (the
reactant/product for which the masses are to be calculated, O in this example). 2
Thus 2 moles of the known, KClO , produces 3 moles of the unknown, 3 O. 2
or 1 mole of the known (KClO ) produces 1.5 moles of the unknown (O ) 3 2
IX - 1
From Topic 7, the conversion of moles to mass for any species is achieved simply by multiplying the number of moles by the molar mass (gram formula weight) of that species, as amount in moles = mass in grams molar mass or mass in grams = amount in moles × molar mass
Amount in moles of KClO 3 in 3.00 g = mass KClO 3 = 3.00 = 0.0245 mole molar mass 122. Therefore, amount in moles of O 2 produced = 1.5 × 0.0245 = 0.0368 mole
and mass of O 2 produced = molar mass of O 2 × moles O 2
= 32.00 × 0.0368 = 1.18 g
In the above, for convenience the term “molecules” was applied to ionic species such as NaOH which do not actually exist as molecules. Strictly the more cumbersome term such as “formula units” would be more correct. Also note that in this example, the moles of unknown produced from 1 mole of the known was deduced as an additional step (shown in bold above). This equation factor can then be used directly to multiply the actual moles of the known to find the moles of the unknown. This procedure should always be followed whenever the stoichiometric coefficient of the known in the reaction equation is not already 1.
Example 2. What mass of carbon dioxide and of water is produced from combustion of 10.0 g of lighter fluid, butane (C H 4 10 ), and what mass of oxygen is consumed in the reaction? Again start by writing a balanced formula equation for the reaction.
2C H 4 10 + 13O 2 v 8CO 2 + 10H O 2
Then write down the stoichiometric coefficients.
2 molecules + 13 molecules v 8 molecules + 10 molecules
2 moles + 13 moles v 8 moles + 10 moles
Using the relationship mass in grams = amount in moles × molar mass gives
2 × 58.1 g + 13 × 32.0 g v 8 × 44.0 g + 10 × 18.0 g
116.2 g + 416.0 g v 352.0 g + 180.0 g
Thence deduce amounts associated with the reaction of 1.0 g of butane:
1.0 g + 416.0/116.2 = 3.6 g v 352.0/116.2 = 3.0 g + 180.0/116.2 = 1.6 g
and thus amounts associated with 10.0 g of butane:
10.0 g + 36.0 g v 30.0 g + 16.0 g
à 10.0 g of butane uses 36.0 g of oxygen and forms 30.0 g of carbon dioxide and 16.
g of water.
Example 5. “limiting reagent” problems. (i) Samples of iron and sulfur, each having a mass of 3.00 g, are heated together to produce iron(II) sulfide until there is no further reaction. Which reactant is in excess and by what mass?
Equation: Fe + S v FeS Ratio: 1 mole 1 mole Moles of Fe available = mass / molar mass = 3.00 / 55.9 = 0.0537 mole Moles of S available = mass / molar mass = 3.00 / 32.07 = 0.0936 mole From the ratios above, 1.00 mole of Fe requires 1.00 mole of S for complete reaction. Moles of S available = 0.0936 which would react with 1.00 × 0.0537 mole of Fe. Amount of Fe available = 0.0537 mole, therefore S is in excess by 0.0936! 0.0537 = 0.0399 mole of S. Mass of S in excess = moles × molar mass = 0.0399 × 32.07 g = 1.28 g.
(ii) A piece of iron (5.59 g) is ignited in a vessel containing 1.60 g of oxygen gas to form Fe O. 3 4 Deduce which reactant is in excess and calculate the amount in moles by which it is in excess.
Equation: 3Fe + 2O 2 v Fe O 3 4 Ratio: 3 mole 2 mole Moles of Fe available = mass / molar mass = 5.59 / 55.9 = 0.100 mole Moles of O 2 available = mass / molar mass = 1.60 / 32.0 = 0.0500 mole From the ratios above, 1 mole of O 2 requires 1.5 mole of Fe for complete reaction. Moles of O 2 available = 0.0500 which would react with 1.50 × 0.0500 mole of Fe = 0.0750 mole Fe Amount of Fe available = 0.100 mole, therefore it is in excess by 0.025 mole
In these examples, a slight variation was introduced in which a known amount of both species was given instead of just one known amount. As can be seen from the solution, essentially the same procedure applies as in the earlier examples, except in the final stage of the calculation when either of the species can be chosen as the known and the amount of the other that would be required for quantitative conversion is calculated and then compared with the actual amount available. Note again there was no need to calculate the amount of the product as this was not requested in the question.
Percentage yield calculations. In all the preceding examples, it has been assumed that complete conversion of reactants to products has occurred (quantitative conversion). The amount of product
obtained when there is 100% conversion is called the THEORETICAL YIELD. For various reasons, conversion in reactions may not be complete or practical techniques used for separating and purifying materials may result in some loss of that material. Consequently, the amount of product obtained in practice, called the ACTUAL YIELD , is usually less than the theoretical yield. The usual way of expressing the yield obtained is as the PERCENTAGE YIELD.
The percentage yield is given by the expression
% yield = actual yield × 100 theoretical yield
Note that the units used for expressing the actual and theoretical yields are usually mass units, but they can be expressed as either mass or moles, provided that the same units are used for both.
Example 6. Calcium (50.5 g) is heated with excess nitrogen to form calcium nitride (60.2 g). Calculate the percentage yield obtained.
Equation: 3Ca + N 2 v Ca N 3 2
Ratios: 3 moles 1 mole
Whence 3 moles of the known, Ca, forms 1 mole of the unknown, Ca N. 3 2
à 1 mole of the known, Ca, forms 1/3 mole of the unknown, Ca N. 3 2
Amount of Ca = mass / molar mass = 50.5 / 40.08 = 1.26 mole
For 100% conversion, amount of Ca N 3 2 produced = 1/3 × 1.26 mole = 0.420 mole
Mass of 0.420 mole Ca N 3 2 = moles × molar mass
= 0.420 × 148.26 g = 62.3 g (theoretical yield)
Actual yield = 60.2 g
Therefore percentage yield = actual yield × 100 = 60.2 × 100 = 96.6 % theoretical yield 62.
Example 7. Cobalt(II) chloride (12.99 g) is converted to cobalt(II) sulfate (14.90 g) by a series of reactions. Calculate the percentage yield obtained. [Insufficient information has been given to write an equation for the conversion steps, but the full equation is not needed - all that is required is the formula for the two compounds, cobalt(II) chloride and cobalt(II) sulfate.] 1 mole of cobalt(II) chloride, CoCl , contains 1 mole of cobalt(II) ions. 2 1 mole of cobalt(II) sulfate, CoSO , contains 1 mole of cobalt(II) ions. 4 Therefore complete conversion of 1 mole of CoCl 2 to CoSO 4 would produce 1 mole of
CoSO 4
Moles of CoCl 2 used = mass / molar mass = 12.99 / 129.9 = 0.100 mole
TUTORIAL QUESTIONS - TOPIC 9.
- For each of the reactions represented by the following equations, supply the number of moles and masses for all of the reactants and products, assuming complete reaction in which 1.0 mole of the specified reactant is used or 1.0 mole of the specified product is produced. Quote your answers to the nearest gram of reactants and products. Part (a) is an example:
(a) For 1.0 mole of SO 2 reacting,
Equation: O (g) 2 + 2SO (g) 2 v 2SO (g) 3 à Mole ratios of 9 9 9
reactants/products: 1 2 2 [from the equation coefficients]
Moles required: 0.50 mole 1.0 mole 1.0 mole [from the question data and the above ratios] Molar masses: 32.0 g mol –1^ 64.1 g mol –1^ 80.1 g mol –1[from atomic weight data]
Masses required: 0.50 × 32.0 g 1.0 × 64.1 g 1.0 × 80.1 g [moles × molar mass] = 16 g = 64 g = 80 g
(b) For 1.0 mole of CaCO 3 reacting,
Equation: CaCO 3 v CaO + CO 2
Mole ratios:
Moles required: 1.0 mole
Molar masses:
Masses required:
(c) For 1.0 mole of Fe reacting
Equation: 2Fe + 3Cl 2 v 2FeCl 3
Mole ratios:
Moles required: 1.0 mole
Molar masses:
Masses required:
(d) For 1.0 mole of H O 2 2 reacting, Equation: 2H O 2 2 v 2H O 2 + O 2
Mole ratios:
Moles required: 1.0 mole
Molar masses:
Masses required:
(e) For 1.0 mole of O 2 produced, Equation: 2Pb(NO ) 3 2 v 2PbO + O 2 + 4NO 2
Mole ratios:
Moles required: 1.0 mole
Molar masses:
Masses required:
(f) For 1.0 mole of N 2 reacting,
Equation: N 2 + 3H 2 v 2NH 3
Mole ratios:
Moles required: 1.0 mole
Molar masses:
Masses required:
(g) For 1.0 mole of H 2 reacting, Equation: N 2 + 3H 2 v 2NH 3
Mole ratios:
Moles required: 1.0 mole
Molar masses:
Masses required:
(h) For 1.0 mole of NH 3 produced, Equation: N 2 + 3H 2 v 2NH 3
Mole ratios:
Moles required: 1.0 mole
Molar masses:
Masses required:
(n) For 1.0 mole of C H 6 14 reacting,
Equation: 2C H 6 14 + 19O 2 v 12CO 2 + 14H O 2
Mole ratios:
Moles required: 1.0 mole
Molar masses:
Masses required:
(o) For 1.0 mole of CO 2 produced,
Equation: 2C H 6 14 + 19O 2 v 12CO 2 + 14H O 2
Mole ratios:
Moles required: 1.0 mole
Molar masses:
Masses required:
(p) For 1.0 mole of Li N produced, 3
Equation: 6Li + N 2 v 2Li N 3
Mole ratios:
Moles required: 1.0 mole
Molar masses:
Masses required:
(q) For 1.0 mole of Cl 2 produced,
Equation: 4HCl + O 2 v 2H O 2 + 2Cl 2
Mole ratios:
Moles required: 1.0 mole
Molar masses:
Masses required:
- Hydrogen gas combines with chlorine gas to form hydrogen chloride. If 2.00 mole of hydrogen reacts with excess chlorine, what mass of hydrogen chloride can be obtained?
- What mass of carbon dioxide is produced when 18 g of carbon is burnt in excess oxygen?
- Hydrogen and chlorine gases combine to form hydrogen chloride gas, HCl. Calculate the minimum mass of hydrogen gas and of chlorine gas needed to produce 10.0 g of hydrogen chloride.
- Calculate the mass of chlorine needed to produce 5.85 g of sodium chloride when excess sodium is heated with chlorine gas.
- Iron burns in air to form the compound Fe O. 3 4
(a) Write the equation for the reaction.
(b) How many moles of oxygen gas are needed to burn 1.0 mole of iron?
(c) What is the mass of this amount of oxygen gas?
(d) Would 0.050 mole of oxygen be enough to cause 5.6 g of iron to react completely to form Fe O? 3 4
- Aluminium sulfide, Al S , can be prepared by reacting aluminium with sulfur. 2 3 (i) What mass of aluminium sulfide can be prepared from 10.0 g of aluminium mixed with 15.0 g of sulfur?
(ii) Which is the non-limiting reagent and by what mass is it in excess?
- Sulfur dioxide gas can be obtained by heating solid sodium hydrogensulfite with sulfuric acid as shown in the following equation:
2NaHSO 3 + H SO 2 4 v 2SO 2 + 2H O + Na SO 2 2 4
(i) Industrially, instead of using moles as gram formula weights for the molar masses of reactants and products, often units of tonne formula weights are preferred. How many tonnes of sulfur dioxide gas can be obtained from 1.00 tonne of sodium hydrogensulfite?
(ii) The standard glob was a unit of mass used by ancient Martians before they succumbed to the loss of atmosphere through the greenhouse effect. How many globs of SO 2 can be prepared from 5.0 globs of NaHSO 3 in the reaction represented by the above equation?
- Calculate the mass of sodium hydroxide required to react exactly with 0.50 mole of nitric acid, HNO. 3
- Nickel(II) nitrate-6-water has the formula Ni(NO ) .6H O. If 29.1 g of this salt 3 2 2 were dissolved completely in water, calculate the following for the resulting solution: (a) The number of moles of Ni 2+ions present.
(b) The number of moles of NO 3 – ions present.
(c) The number of nickel ions present.
(d) The number of nitrate ions present.
- A piece of pure zinc weighing 15.8 g is dissolved completely in hydrochloric acid. Calculate:
(a) the mass of hydrogen gas formed
(b) the mass of zinc chloride that could be formed by evaporating the water from the resultant solution.
- Self Help Problems modules “Mole Concept 1" Q 12, 13, 14, 15(i). “Mole Concept 2" Q1.
ANSWERS TO TUTORIAL TOPIC 9
- (b) For 1.0 mole of CaCO 3 reacting, Equation: CaCO 3 v CaO + CO 2
Mole ratios: 1 1 1
Moles required: 1.0 1.0 1.
Molar masses: 100.1 g mol –1 56.1 g mol –1^ 44.0 g mol–
Masses required: 1.0 × 100.1 1.0 × 56.1 1.0 × 44.
= 100 g = 56 g = 44 g
(h) For 1.0 mole of NH 3 produced, Equation: N 2 + 3H 2 v 2NH 3 Mole ratios: 1 3 2 Moles required: 0.50 1.5 1. Molar masses: 28.0 g mol –1^ 2.0 g mol –1^ 17.0 g mol– Masses required: 0.50 × 28.0 1.5 × 2.0 1.0 × 17. = 14 g = 3 g = 17 g
(i) For 1.0 mole of Pb(NO ) 3 2 reacting Equation: 2Pb(NO ) 3 2 v 2PbO + O 2 + 4NO 2 Mole ratios: 2 2 1 4 Moles required: 1.0 1.0 0.50 2. Molar masses: 331 g mol –1^ 223 g mol –1^ 32 g mol –1^ 46 g mol– Masses required: 1.0 × 331 1.0 × 223 0.50 × 32 2.0 × 46 = 331 g = 223 g = 16 g = 92 g
(j) For 1.0 mole of ZnS reacting, Equation: 2ZnS + 3O 2 v 2ZnO + 2SO 2 Mole ratios: 2 3 2 2 Moles required: 1.0 1.5 1.0 1. Molar masses: 97.5 g mol –1^ 32.0 g mol –1^ 81.4 g mol –1^ 64.1 g mol– Masses required: 1.0 × 97.5 1.5 × 32.0 1.0 × 81.4 1.0 × 64. = 97.5 g = 48.0 g = 81.4 g = 64.1 g = 97 g = 48 g = 81 g = 64 g
(k) For 1.0 mole of O 2 reacting, Equation: 2ZnS + 3O 2 v 2ZnO + 2SO 2 Mole ratios: 2 3 2 2 Moles required: 2/3 = 0.667 1.0 2/3 = 0.667 2/3 = 0. Molar masses: 97.5 g mol –1^ 32.0 g mol –1^ 81.4 g mol –1^ 64.1 g mol– Masses required: 0.667 × 97.5 1.0 × 32.0 0.667 × 81.4 0.667 × 64. = 65.0 g = 32.0 g = 54.3 g = 42.7 g = 65 g = 32 g = 54 g = 43 g
(l) For 1.0 mole of NaHCO 3 reacting, Equation: 2NaHCO 3 v Na CO 2 3 + CO 2 + H O 2 Mole ratios: 2 1 1 1 Moles required: 1.0 0.50 0.50 0. Molar masses: 84.0 g mol –1^ 106.0 g mol –1^ 44.0 g mol –1^ 18.0 g mol– Masses required: 1.0 × 84.0 0.50 × 106.0 0.50 × 44.0 0.50 × 18. = 84 g = 53 g = 22 g = 9 g
(m) For 1.0 mole of C H 2 6 reacting, Equation: 2C H 2 6 + 7O 2 v 4CO 2 + 6H O 2 Mole ratios: 2 7 4 6 Moles required: 1.0 3.5 2.0 3. Molar masses: 30.1 g mol –1^ 32.0 g mol –1^ 44.0 g mol –1^ 18.0 g mol– Masses required: 1.0 × 30.1 3.5 × 32.0 2.0 × 44.0 3.0 × 18. = 30 g = 112 g = 88 g = 54 g
(n) For 1.0 mole of C H 6 14 reacting, Equation: 2C H 6 14 + 19O 2 v 12CO 2 + 14H O 2 Mole ratios: 2 19 12 14 Moles required: 1.0 9.5 6.0 7. Molar masses: 86.2 g mol –1^ 32.0 g mol –1^ 44.0 g mol –1^ 18.0 g mol– Masses required: 1.0 × 86.2 9.5 × 32.0 6.0 × 44.0 7.0 × 18. = 86 g = 304 g = 264 g = 126 g
(o) For 1.0 mole of CO 2 produced, Equation: 2C H 6 14 + 19O 2 v 12CO 2 + 14H O 2 Mole ratios: 2 19 12 14 Moles required: 2/12 = 0.167 19/12 = 1.58 1.0 14/12 = 1. Molar masses: 86.2 g mol –1^ 32.0 g mol –1^ 44.0 g mol –1^ 18.0 g mol– Masses required: 0.167 × 86.2 1.58 × 32.0 1.0 × 44.0 1.17 × 18. = 14.4 g = 50.6 g = 44.0 g = 21.0 g = 14 g = 51 g = 44 g = 21 g
(p) For 1.0 mole of Li N produced, 3 Equation: 6Li + N 2 v 2Li N 3 Mole ratios: 6 1 2 Moles required: 3.0 0.50 1. Molar masses: 6.9 g mol –1^ 28.0 g mol –1^ 34.8 g mol– Masses required: 3.0 × 6.9 0.50 × 28.0 1.0 × 34. = 21 g = 14 g = 35 g
(q) For 1.0 mole of Cl 2 produced, Equation: 4HCl + O 2 v 2H O 2 + 2Cl 2 Mole ratios: 4 1 2 2 Moles required: 2.0 0.50 1.0 1. Molar masses: 36.5 g mol –1^ 32.0 g mol –1^ 18.0 g mol –1^ 70.9 g mol– Masses required: 2.0 × 36.5 0.5 × 32.0 1.0 × 18.0 1.0 × 70. = 73 g = 16 g = 18 g = 71 g
à moles of C available = 18 / 12.01 = 1.5 mol
From the mole ratios above, moles of CO 2 formed = 1.5 mol
Mass of carbon dioxide = moles × molar mass of CO 2
= 1.5 × (12.01 + 2 × 16.00) = 66 g
- Balanced equation: H 2 + Cl 2 v 2HCl
Mole ratios: 1 mol 1 mol 2 mol
Mass of hydrogen chloride = 10.0 g
à moles of HCl = mass / molar mass of HCl = 10.0 / (1.01 + 35.45) = 0.274 mol
1 mole of HCl requires ½ mole of H 2 and ½ mole of Cl 2
à moles of H 2 = moles of Cl 2 = 0.500 × 0.274 mol = 0.137 mol
Mass of hydrogen = moles × molar mass of H 2 = 0.137 × 2.02 g = 0.277 g
Mass of chlorine = moles × molar mass of Cl 2 = 0.137 × 70.90 g = 9.71 g
- Balanced Equation: 2Na + Cl 2 v 2NaCl
Mole ratios: 1 mol 2 mol
Mass of sodium chloride = 5.85 g
à moles of NaCl = 5.85 / molar mass of NaCl = 5.85 / 58.5 = 0.100 mol
Moles of Cl 2 required = ½ × moles of NaCl = 0.500 × 0.100 = 0.0500 mol
Mass of Cl 2 = moles × molar mass of Cl 2 = 0.0500 × 70.90 g = 3.55 g
- (a) 3Fe + 2O 2 v Fe O 3 4
(b) From the stoichiometric coefficients in the equation, 3 moles of Fe require 2
moles of O 2 so 1.0 mole of Fe requires 2/3 moles of O 2 = 0.667 mol
(c) Molar mass of O 2 = 2 × 16.00 g = 32.00 g mol–
à 0.667 mole of O 2 has a mass = 0.667 × 32.00 = 21 g
(d) From the equation, 1 mole of O 2 requires 1.5 moles of Fe for complete reaction. à 0.050 mole of O 2 would require 0.050 × 1.5 = 0.075 mole of Fe. Mass of 0.075 mole of Fe = 0.075 × molar mass of Fe = 0.075 × 55.85 g = 4.2 g. Thus there would be an excess of iron = 5.6 – 4.2 = 1.4 g Alternative method: Moles of Fe = mass / molar mass Fe = 5.6 / 55.85 = 0.10 mol From the equation, 1 mole of Fe requires 2/3 mole of O 2 = 0.667 mol à moles of O 2 needed to react with 0.10 mole of Fe = 0.0667 mole, so 0.050 mole of O 2 is insufficient.
- (i) From the given equation, 1 mole NH Cl reacting leads to the formation of 1 mole N O. 4 2 Moles of NH Cl = mass / molar mass = 100.0 / 53.49 = 1.870 mol 4 à moles of N O produced = 1.870 mol 2 and mass of N O = moles × molar mass = 1.870 × 44.01 g = 82.28 g 2
(ii) Theoretical yield = 82.30 g Actual yield = 75.4 g % yield = actual yield × 100 = (75.4 / 82.3) × 100 = 91.6 % theoretical yield
- Equation: Fe O 3 4 + 4H 2 v 3Fe + 4H O 2 Mole ratios: 1 mol 3 mol à 1 mole of Fe requires 1/3 mole of Fe O 3 4 = 0.333 mol. Moles of Fe = mass / molar mass of Fe = 112 / 55.85 = 2.00 mol Moles of Fe O 3 4 = 0.333 × 2.00 = 0.667 mol. Mass of Fe O 3 4 = moles × molar mass of Fe O 3 4 = 0.667 × 231.5 = 154 g
- (i) Equation: 2NaNO 3 v 2NaNO 2 + O 2 Mole ratios: 2 mol 1 mol 1 mole of O 2 requires 2 moles of NaNO 3 Moles of O 2 = mass / molar mass of O 2 = 64.0 / 32.0 = 2.00 mol à moles of NaNO 3 = 2 × 2.00 = 4.00 mol and mass of sodium nitrate = moles × molar mass of NaNO 3 = 4.00 × 85.0 = 340 g
(ii) Let x = mass of sodium nitrate needed for 100% yield. 340 g sodium nitrate produces 80% yield and x g of sodium nitrate produces 100% yield à340/x = 80/ x = 425 g
- (i) Equation: C + O 2 v CO 2 Mole ratios: 1 mol 1 mol 1 mol [Note that this is a limiting reagent type of problem where both reactants have a specified mass and one of them is in excess. The first task is to find out which is the limiting reagent and this will determine how much carbon dioxide can form.] Moles of C = mass / molar mass of C = 1.20 / 12.01 = 0.100 mol Moles of O 2 = mass / molar mass of O 2 = 8.00 / 32.00 = 0.250 mol From the equation, 1 mole of C uses 1 mole of O 2 to produce 1 mole of CO. 2 If all the carbon (0.100 mole) were used it would react with 0.100 mole of O , leaving 2 (0.250 – 0.100) mole of O 2 in excess unreacted and form 0.100 mole of CO , so the 2 carbon is the limiting reagent. Mass of carbon dioxide formed = moles × molar mass of CO 2 = 0.100 × 44.0 = 4.40 g.
