# Molar Ratios, Introduction to Chemistry ## Definition of molar ratio

Chemical equations are symbolic representations of chemical reactions. In a chemical equation, the reacting materials are written on the left, and the products are written on the right; the two sides are usually separated by an arrow showing the direction of the reaction. The numerical coefficient next to each entity denotes the absolute stoichiometric amount used in the reaction. Because the law of conservation of mass dictates that the quantity of each element must remain unchanged over the course of a chemical reaction, each side of a balanced chemical equation must have the same quantity of each particular element.

In a balanced chemical equation, the coefficients can be used to determine the relative amount of molecules, formula units, or moles of compounds that participate in the reaction. The coefficients in a balanced equation can be used as molar ratios, which can act as conversion factors to relate the reactants to the products. These conversion factors state the ratio of reactants that react but do not tell exactly how much of each substance is actually involved in the reaction.

## Determining Molar Ratios

The molar ratios identify how many moles of product are formed from a certain amount of reactant, as well as the number of moles of a reactant needed to completely react with a certain amount of another reactant. For example, look at this equation:

[latex]CH_ <4>+ 2O_ <2>rightarrow CO_ <2>+ 2H_<2>O[/latex]

From this reaction equation, it is possible to deduce the following molar ratios:

In other words, 1 mol of methane will produced 1 mole of carbon dioxide (as long as the reaction goes to completion and there is plenty of oxygen present). These molar ratios can also be expressed as fractions. For example, 1 mol CH4: 1 mol CO2 can be expressed as [latex]frac<1 mol CH_4><1 mol CO_2>[/latex]. These molar ratios will be very important for quantitative chemistry calculations that will be discussed in later concepts.

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## Molar Definition in Chemistry (Unit) • Ph.D., Biomedical Sciences, University of Tennessee at Knoxville
• B.A., Physics and Mathematics, Hastings College

Molar refers to the unit of concentration molarity, which is equal to the number of moles per liter of a solution. In chemistry, the term most often refers to molar concentration of a solute in a solution. Molar concentration has the units mol/L or M.
Molar also refers to other measurements dealing with moles such as molar mass, molar heat capacity and molar volume.

## Example

A 6 molar (6 M) solution of H2SO4 refers to a solution with six moles of sulfuric acid per liter of solution. Remember, the volume refers to liters of solution, not liters of water added to prepare the solution.

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## What Is a Mole Ratio in Chemistry? • Ph.D., Biomedical Sciences, University of Tennessee at Knoxville
• B.A., Physics and Mathematics, Hastings College

In a chemical reaction, compounds react in a set ratio. If the ratio is unbalanced, there will be leftover reactant. To understand this, you need to be familiar with the molar ratio or mole ratio.

## Mole Ratio Definition

A mole ratio is ​the ratio between the amounts in moles of any two compounds involved in a chemical reaction. Mole ratios are used as conversion factors between products and reactants in many chemistry problems. The mole ratio may be determined by examining the coefficients in front of formulas in a balanced chemical equation.

Also known as: The mole ratio is also called the mole-to-mole ratio.

## Mole Ratio Example: Balanced Equation

The mole ratio between O2 and H2O is 1:2. For every 1 mole of O2 used, 2 moles of H2O are formed.

The mole ratio between H2 and H2O is 1:1. For every 2 moles of H2 used, 2 moles of H2O are formed. If 4 moles of hydrogen were used, then 4 moles of water would be produced.

## Unbalanced Equation Example

By inspection, you can see this equation is not balanced because mass is not conserved. There are more oxygen atoms in ozone (O3) than there are in oxygen gas (O2). You cannot calculate mole ratio for an unbalanced equation. Balancing this equation yields:

Now you can use the coefficients in front of ozone and oxygen to find the mole ratio. The ratio is 2 ozone to 3 oxygen, or 2:3. How do you use this? Let’s say you are asked to find how many grams of oxygen are produced when you react 0.2 grams of ozone.

1. The first step is to find how many moles of ozone are in 0.2 grams. (Remember, it’s a molar ratio, so in most equations, the ratio is not the same for grams.)
2. To convert grams to moles, look up the atomic weight of oxygen on the periodic table. There are 16.00 grams of oxygen per mole.
3. To find how many moles there are in 0.2 grams, solve for:
x moles = 0.2 grams * (1 mole/16.00 grams).
You get 0.0125 moles.
4. Use the mole ratio to find how many moles of oxygen are produced by 0.0125 moles of ozone:
moles of oxygen = 0.0125 moles ozone * (3 moles oxygen/2 moles ozone).
Solving for this, you get 0.01875 moles of oxygen gas.
5. Finally, convert​ the number of moles of oxygen gas into grams for the answer:
grams of oxygen gas = 0.01875 moles * (16.00 grams/mole)
grams of oxygen gas = 0.3 grams

It should be fairly obvious that you could have plugged in the mole fraction right away in this particular example because only one type of atom was present on both sides of the equation. However, it’s good to know the procedure for when you come across more complicated problems to solve.

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With contragravity, of course, terms like «escape-velocity» and «mass-ratio» were of purely antiquarian interest.

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## Transesterification processes for biodiesel production from oils and fats

Jan C.J. Bart, . Stefano Cavallaro, in Biodiesel Science and Technology , 2010

### 7.2.3 Molar ratio of alcohol to oil

The molar ratio of alcohol to oil is one of the most important variables affecting the yield of alkyl esters. Stoichiometrically, three moles of alcohol are required per mole of triglyceride, but in practice a (large) excess of alcohol is employed in order to displace the equilibrium to greater ester production. Yield is a most important factor affecting commercial profitability. It is therefore crucial to achieve a near 100% transesterification of TGs and FFAs. Conversions lower than 90% are not of commercial interest. Figure 7.1 shows the effect of alcohol-to-oil ratio on product composition for transesterification [ 6 ]. 7.1 . Effect of alcohol-to-oil ratio on product composition for transesterification (after ref.  ).

Typically, the optimal ratios for methanolysis of Cynara oil are between 4.05 : 1 and 5.67 : 1 [ 44 ]. Below 4.05 : 1 the reaction is reported to be incomplete, whereas above 5.67 : 1 it becomes difficult to separate glycerol from methanol as a by-product. The optimum molar ratio between alcohol and oil ensuring highest reaction rate depends on the type of catalyst used: typically 6 : 1 for alkali-catalysed transesterifications [ 45 , 46 ] and up to 30 : 1 for acid-catalysed alcoholysis [ 6 ]. Even higher molar ratios (up to 45 : 1) may be necessary when the oil contains a large amount of free fatty acids [ 6 ]. In industrial processes a methyl ester yield of over 98% is normally achieved for 100% excess of methanol. Experiments conducted with molar ratios of 8 : 1 and 6 : 1 for mixing speeds of 0–3600 rpm and a catalyst (NaOH) concentration of 0.1 wt% have shown considerably higher conversions of SBO at the molar ratio of 8 : 1 for all mixing levels [ 47 ]. Increasing the molar ratio of alcohol-to-TG increases alcohol recovery and product separation costs. Higher molar ratios also interfere in the separation of glycerol.

Transesterification of RSO in supercritical methanol requires a molar ratio of 42 : 1 to achieve a yield of 95% of methyl esters [ 8 ]. One-phase methanolysis of SBO in the presence of a co-solvent benefits from an alcohol-to-oil ratio of 27 : 1 ( Fig. 7.2 ). The transesterification reaction is also shifted to the right by removal of one of the products from the reaction mixture. This is a much preferred option to drive the reaction to completion. 7.2 . Effect of alcohol/vegetable oil ratio for one-phase methanolysis of soybean oil at 296 K (replotted from data in ref.  ).

## Dry (CO2) Reforming

### 7.2.2.1 Effect of CO2/Fuel Ratio

The stoichiometric molar ratio of CO 2/CH4 is 1 in dry reforming of CH4 (Eq. 7.1 ). Excess amount of CO2 in feed could increase the conversion of methane. Figure 7.1 shows the plot of CO2/CH4 feed ratio versus equilibrium amount of product species at 750 °C and 900 °C. Both CO and H2 are major products in a wide range of CO2/CH4 ratios at elevated temperatures; feeding CO2/CH4 at a ratio of 1.0 mainly yields CO and H2. Higher CO yield could be achieved with rising CO2/CH4 ratio, but the yield of H2 decreases as the formation of byproduct H2O via the reverse water-gas shift reaction (Eq. 7.4 ). FIGURE 7.1 . Effect of CO2/CH4 ratio on dry reforming of methane; equilibrium amount of product species from 1 mole of methane at 750 °C (a) and 900 °C (b), 0.1 MPa (The amount of carbon (C) and CH4 formed are not shown for 900 °C (b) since their concentrations in the reaction product are low).

## Dyeing with insoluble azoic colour

### 11.8.2 Diazotization and coupling

#### NaNO2

It produces HNO2 after reaction with HCl, which in turn diazotizes the base.

It reacts with NaNO2 to produce HNO2 as well as converts free bases into water soluble salts

This hydrochloride of base reacts with HNO2 to produce diazonium salt.

H2SO4 is not preferred for diazotization because sulphates are less soluble than chlorides.

Generally, a molar ratio of base, NaNO 2 and HCl is maintained at 1: 1: 3. One molecule of base requires one molecule of HNO2 for diazotization which comes from one molecule of NaNO2. Out of 2.5 molecules of HCl, one forms base hydrochloride, one reacts with NaNO2 to produce HNO2 while the half maintains pH during diazotization.

Diazotization reaction is an exothermic process; ice is added to maintain temperature (0–5°C). If ice is not added, especially above 30°C, diazotized salt decomposes, react with undiazotized salt and water resulting loss of a part of base.

After diazotization, the solution has a strong acidic pH (2–3) due to presence of excess HCl; for efficient coupling at pH

5–6, the pH is raised by adding CH3COONa (3–4%), which converts inorganic HCl to organic CH3COOH, thus reducing acidity of bath.

CH3COONa and CH3COOH combination act as buffer to maintain desired pH; however, some extra CH3COOH needs to be added to bath to maintain pH.

#### Alkali binding agent

Naphthol treated fabric before development has strong alkaline pH and the coupling bath has a pH

5–6. When naphtholated fabric is dipped in coupling bath for reaction, alkali attached with naphthol becomes free and migrates to the bath, neutralizes acid and coupling is either retarded or stopped. Application of Al2(SO4)3 or more commonly used CH3COOH as alkali binding agent neutralizes this NaOH and maintains acidity of bath required for coupling.

It promotes aggregation of dye molecules within fibre and removes superficial pigments.

## Edible and Nonedible Biodiesel Feedstocks

### 17.6.2 Esterification Process

In this process, the molar ratio of methanol to crude oils with high acid values was maintained at 12:1 (50% v/v); 1% (v/v) of sulfuric acid (H 2SO4) was added to the preheated oils at 60°C for 3 h and 400 rpm stirring speed in a glass reactor. On completion of this reaction, the products were poured into a separating funnel to separate the excess alcohol, sulfuric acid, and impurities presented in the upper layer. The lower layer was separated and entered into a rotary evaporator and heated at 95°C under vacuum conditions for 1 h to remove methanol and water from the esterified oil.

Due to the high acid value of some crude oils, the amount of methanol can be increased to reduce the acid value to less than 4 mg KOH/g oil.

## Lipase-Catalyzed Biodiesel Production

### Solvents

In the above section ( molar ratio ) discussed that excess amount of alcohols increases FAME yield. In order to increase the solubility of alcohol (not the enzyme), solvents are used and they alleviate negative effect of methanol on the catalyst and precede the transesterification. Enzyme should be insoluble in solvent; otherwise, it will not be active ( Kanerva et al., 1990 ; Antczak et al., 2009 ). Various hydrophilic and hydrophobic organic solvents such as cyclohexane, n-hexane, tert-butanol, petroleum ether, isooctane and 1,4-dioxane are mainly studied organic solvents in enzymatic biodiesel production. If organic solvent is used as medium, overall alcohol is added at the beginning of the reaction. In solvent-free reaction medium, alcohol is added stepwise to prevent enzyme activity with high alcohol concentration ( Sevil et al., 2012 ).

## Natural Gas Conversion V

Jordi Llorca, . Pilar Ramírez de la Piscina, in Studies in Surface Science and Catalysis , 1998

### 2 EXPERIMENTAL

Bimetallic catalysts with a nominal molar ratio Pt:Sn = 1:1 were prepared by anchoring the cis-[PtCl(SnCl3) (PPh3)2] complex from a methylene chloride solution onto γ-A12O3 (Girdler; surface area of 188 m 2 g — 1 ), MgO (obtained by adding ammonia to a MgCl2 solution; 110 m 2 g — 1 ) and SiO2 (Degussa aerosil; 200 m 2 g — 1 ) at room temperature followed by a vacuum treatment at 373 K overnight, as described elsewhere  . All supports were partially dehydrated by treatment under high vacuum at 473 K for 16 h. Samples were reduced in a glass reactor at atmospheric pressure in flowing hydrogen (40 ml min — 1 ) at 673 K for 16 h. The catalysts obtained are referred to as PtSn/Al2O3, PtSn/MgO and PtSn/SiO2. For comparative purposes, platinum catalysts were also prepared in a similar manner by anchoring the cis-[PtCl2,(PPh3)2] complex onto the supports. The resulting samples are labelled Pt/Al2O3, Pt/MgO and Pt/SiO2. All samples were prepared without contact with air. Table 1 shows the metal content of catalysts.

Table 1 . Composition and initial catalytic activity in the n-hexane transformations of platinum based catalysts.

Catalyst Pt (%) Sn (%) Initial activity (μmol.g — 1 Pt.h — 1 ) a
Pt/Al2O3 2.37 1097
Pt/MgO 2.78 1071
Pt/SiO2 1.96 1214
PtSn/Al2O3 2.79 1.81 799
PtSn/MgO 2.66 1.76 594
PtSn/SiO2 2.52 1.55 278

The n-hexane reaction was studied in a continuous-flow glass microreactor operating at atmospheric pressure. A feed of hydrogen saturated with n-hexane of varying composition was generated by bubbling hydrogen through a thermostatted saturator. 20-40 mg of catalyst diluted with inactive SiC was used for each reaction test. Products were separated in a TRB-1 capillary column on a gas chromatograph equipped with an automated gas sample valve. 3-hexenes were not quantified due to their low yield and proximity to the n- hexane peak. Prior to reaction studies, all catalysts were re-reduced in situ at 673 K for 1 h in flowing hydrogen. After reduction, the reactor temperature was lowered to 453 K and the catalyst was subsequently aged by increasing the temperature from 453 to 753 K at 5 K min — 1 in a reaction mixture of H2:n- hexane = 4:1 (20 ml min — 1 ). The transformation of n-hexane was then monitored for 20-22 h in these reaction conditions for all catalysts. After the accelerated aging treatment, reaction studies were carried out at 753 K with H2:n-hexane molar ratios ranging from 4 to 15 and reagent flow from 5 to 40 ml min — 1 in order to study the performance and stability of catalysts.

## Biodiesel

### Ratio of alcohol to oil

Another important variable affecting the yield of ester is the molar ratio of alcohol to vegetable oil. A molar ratio of 6:1 is normally used in industrial processes to obtain methyl ester yields higher than 98 per cent by weight. Higher molar ratio of alcohol to vegetable oil interferes in the separation of glycerol. It was observed that lower molar ratios required more reaction time. With higher molar ratios, conversion increased but recovery decreased due to poor separation of glycerol. It was found that optimum molar ratios depend upon type and quality of oil.

## Synthesis and Application of Au–Fe3O4 Dumbbell-Like Nanoparticles

### 30.2.3.1 The Molar Ratio of Au and Fe Precursors

Grzybowski et al. discovered that different molar ratio of Au and Fe precursors led to a gradual change of morphology from dumbbell-like to flower-like  . When more Fe precursor was used relative to Au precursor, the formation of nanoflowers was preferred, and much larger flower petals were observed when the relative amount of Fe precursor increased. Fig. 30.4 shows the Au–Fe3O4 nanoparticles prepared by refluxing different amounts of Fe(CO)5 in 1-octadecene at the presence of Au nanoparticles, oleic acid, and oleylamine for 50 min. When the molar ratio of Au and Fe precursors was 1:1 ( Fig. 30.4A ), the nanoparticles are dumbbell-like dimers comprising a smaller (5–8 nm) Fe3O4 domain attached to a larger (10 nm) Au part. The size of the Fe3O4 domain increases with the increase of the molar ratio of Au and Fe precursors and, at 1:2.5, is similar to that of the Au particle ( Fig. 30.4B ). At higher values of Au:Fe, multiple Fe3O4 domains form on each Au nanoparticle. For example, at 1:6 ( Fig. 30.4C ), about 60% of the composite particles have two and about 20% have three Fe3O4 domains around the Au core, while at 1:10 ( Fig. 30.4D ), about 80% of the particles have three or four Fe3O4 leaves per individual Au nanoparticle. Figure 30.4 . TEM images of Au–Fe3O4 nanoparticles prepared in 1-octadecene at the presence of oleic acid and oleylamine with Au:Fe initial molar ratios of (A) 1:1, (B) 1:2.5, (C) 1:6, and (D) 1:10. The dark portions of Au–Fe3O4 nanoparticles correspond to Au while lighter ones to Fe3O4. All scale bars are 20 nm.

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